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Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Let the total mass of the solution be 100g and mass of benzene be 30 g
therefore mass of tetrachloride= (100-30)g = 70g
Molar mass of benzene,
solution;
Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.
(a) Mol. mass of
Moles of
Volume of solution = 4.3 L
Molarity,
(b) Number of moles present in 1000 ml of 0.5M H2SO4= 0.5 mol
therefore number of moles present in 30ml of 0.5M H2SO4=mol =0.015mol
therefore molarity =0.015/0.5L
thus molarity is 0.03M
Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as:
Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.
It is given that the solubility of H2S in water at STP is 0.195m, i.e., 0.195 mol of H2S is dissolved in1000 g of water.
=
At STP pressure (P) = 0.987bar
According to henry's law p = kH x
Solution :
given that,
pressure of CO2=2.5atm
1 atm = 1.01325x 105 pa so that
pressure of CO2 = 2.5x1.01325x105pa
= 2.533125 x 105 pa
KH = 1.67 x 108 pa
ACcording to henry's law p= KH*X
or X=P/KH
= 2.533125 x 105/1.67x 108
= 1.52 x 10-3
But we have 500ML odf soda water so that
Volume of water = 500mL
Density of water =1g/ml
mass = volume x density
500 mL of water = 500g of water
molar mass f water (H2O) = 18g mol-1
number of moles =
Vapoure pressure of pure water (solvent) at 298 K, p0 = 23.8 mm
Vapour pressure of solution, p = ?
Mass of solvent ,W = 850 g
Mass of solute,M = 50 g
Mol. mass of water (H2O), M = 18 g mol–1
Mol.mass of urea NH2 CO NH2
= 14 + 2 + 12 + 16 + 14 + 2
= 60 g mol–1
According to Raoult's law,
Hence, 23.78 mm Hg. Ans.
Solution:
Mass of acetic acid W1= 75g
Molar mass of
lowering of melting point ,
Hence, 5.08 g of ascorbic acid is needed to be dissolved.
Insoluble: toluene, chloroform.
Partially soluble: phenol, pentanol.
Highly soluble: formic acid, ethylene glycol.
Mole fraction: Commonly used symbol for mole fraction is x and subscript used on the right hand side of x denotes the component.
It is defined as:
Mole fraction of a component =
For example, in a binary mixture, if the number of moles of A and B are nA and nB respectively, the mole fraction of A will be
x =
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The most commonly used form of Henry’s law states that “the partial pressure of the gas
in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution”
Why is the elivation in boiling point of water different in the following solution?
(i) 0.1 molar NaCl solution.
(ii) 0.1 molar sugar solution.
Molality: Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as:
Molality (m) =
molality are independent of temperature, whereas molarity is a function of temperature. This is because volume depends on temperature
and the mass does not.
Henry’s law states that “the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution” and is expressed as: p = KHx
where KH henry's constant
van’t Hoff factor, to account for the extent of dissociation or association. This factor i is defined as:
Answer:
The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. The ideal solutions have two other important properties. The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is
also zero, i.e.,
ΔmixH = 0, ΔmixV = 0
Molarity: Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution,
Molality: Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as:
molality are independent of temperature, whereas molarity is a function of temperature. This is because volume depends on temperature and mass does not.
For dilutesolutions the elevation of boiling point (ΔTb) is directly proportional to the molal concentration of the solute in a solution. Thus
ΔTb ∝ m
or ΔTb = Kbm
Here m (molality) is the number of moles of solute dissolved in 1 kg of solvent and the constant of proportionality, Kb is called Boiling Point
Elevation Constant or Molal Elevation Constant . The unit of Kb is K kg mol-1.
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Answer:
A hypotonic solution in which it contain more solute than solvent.example a lot of salt (solute)
dissovled in water (solvent)
isotonic solution in which solute and solvent are equally distrubuted for example a cell normally wants to remain in an isotonic solution where concentration of liquid inside it is equal to the concentration of the liquid outside of it.
hypertonic solution in which contains more solvent than solute.
the elevation of boiling point (ΔTb) is directly proportional to the molal concentration of
the solute in a solution. Thus
ΔTb ∝ m
or ΔTb = Kbm
Here m (molality) is the number of moles of solute dissolved in 1 kg
of solvent and the constant of proportionality, Kb is called Boiling Point
Elevation Constant or Molal Elevation Constant (Ebullioscopic
Constant). The unit of Kb is K kg mol-1.
osmotic pressure is proportional to the molarity, C of the solution at a given temperature T. Thus:
Π = C R T
Here Π is the osmotic pressure and R is the
gas constant.
Π = (n2 /V) R T
Here V is volume of a solution in litres containing n2 moles of solute.
If w2 grams of solute, of molar mass, M2 is present in the solution, then n2 = w2 / M2 and we can write
Thus, knowing the quantities w2, T, Π and V we can calculate the
molar mass of the solute.
p1 ∝ x1
and p1 = x1
where is vapour pressure in the pure state
Answer:
The vapour pressure of a liquid depend on following factor.
(i) Temperature.
(ii) Type of solute in solution (i.e., whether the solute is volatile or non-volatile.
Answer:
There are two types of azetropes:
(i) Minimum boiling azetropes →Ethanol + water.
(ii) Maximum boiling azetropes →Chloroform + acetone.
Answer:
The two characteristics of ideal solution is:
(i) An ideal solution follows Raoult’s law and the components can be separated by fractional distillation.
(ii) ΔHmix = 0,
ΔVmix = 0,
A-A, B-B and A-B interactions of same length.
Answer:
A solution showing negtive deviation:
ΔH = –ve,
ΔV = – ve, Ptotal < P0A XA + P0B XB
Example: Solution of chloroform and acetone.
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Answer:
(i) Plants absorb water from the soil through their roots due to osmosis.
(ii) In animals, water moves into different parts of body due to osmosis.
Answer:
A solution is a homogenous mixture of two or more than two substances whose composition can change within a certain limits. A solution of two substances is called binary solution.
In solution, the component that present in small amount is known as solute and the component present in larger amount known as solvent.
Nine kinds of solution are possible.
(i) Gas in gas. When one gas is mixed with another gas, it is called solution of gas in gas. Example : Air is a mixture of nitrogen and oxygen.
(ii) Liquid in gas. When liquid is mixed with large amount of gas, it is called liquid in gas solution. Example: Moisture (water in air).
(iii) Solid in gas. When small amount of solid particles are dispersed in gas, it is called solution of solid in gas. Example: Smoke.
(iv) Gas in liquid. When gas is dissolved in liquid, it is called gas in liquid solution. Examples: CO2 gas dissolved in water, oxygen dissolved in water.
(v) Liquid in liquid. When a liquid is miscible with another liquid, it forms solution of liquid in liquid. Examples. Ethanol dissolved in water, methanol dissolved in water.
(vi) Solid in liquid. When solid is dissolved in water, the solution is called solid in liquid. Examples: Cane sugar dissolved in water, sodium chloride dissolved in water.
(vii) Gas in solid. When gas is present, the solution is called gas in solid. Example: H2 gas in palladium.
(viii) Liquid in solid. When liquid is present in solid, the homogeneous mixture is called solution of liquid in solid. Example: Amalgam of mercury with sodium.
(ix) Solid in solid. When solid is dissolved in another solid, the homogeneous mixture is called solution of solid in solid. Examples: Alloys are solid in solid solution, copper dissolved in gold.
Answer:
In both the components, water (A component) and alcohol (B component), the molecules are hydrogen bonded, i.e., A-A as well as B-B interactions are mainly H-bonds forces. When these two components are mixed to form the solution, due to molecular collisions A-A and B-B interactions are broken down and new interactions of the type A-B are formed. A-B interactions are also mainly H-bonds i.e., the molecules of one liquid will tend to break the hydrogen bonds in the molecules of other liquid and vice-versa. A-B interactions formed will be weaker than A-A and B-B types of of interactions as the alcohol and water solution show positive deviation and it will lead to increase in volume (ΔVmix is + ve).
Answer:
For most of gases like N2, O2, He etc. the value of Henry's constant, KH increases with temperature and as a result of this solubility of gases decreases with increase of temperature. We know, at given temperature.
Partial pressure of gas in solution = KH x mole fraction of gas in sol. KH depends on the nature of gas. For example, KH value of O2 at 293 K is 34.86 while at 393 K it is 46.82 kilo bar.
Answer:
Henry’s law states that the solubility of a gas in a liquid is directly proportional to pressure of the gas; temperature constant.
Mathematically, P = KH x
where
x=mole fraction of gas in solution
P is partial pressure of gas,
KH is Henry's constant.
Applications:
(1) Henry's law finds various applications in industry and enables us to explain and understand some biological phenomena. The some of important applications are : CO2 solubility in soft drinks, beverages, soda water etc. is increased by applying high pressure and bottles are sealed under high pressure.
(2) For deep divers, oxygen diluted with less soluble He gas is used as breathing gas and it minimises the painful effects due to higher solubility of N2 gas in blood.
(3) In lungs, where oxygen is present in air with high partial pressure, haemoglobin combines with O2 to form oxyhaemoglobin. In tissues where partial pressure of O2 is low, oxyhaemoglobin releases the oxygen for utilisation in cellular activities.
Answer:
According to Raoult's law, the partial vapour pressure of each component in any solution is directly proportional to its mole fraction.
The solution which obeys Raoult's law over enitre range is known as ideal solution.
The solution which do not obeys Raoult's law is known as non- ideal solution.
Non-ideal solution have vapour pressure either higher or lower is predicted by raoult's law
If the vapour pressure is higher then solution is said to exhibit positive deviation, And if the vapour pressure is lower than the solution than it said to be negtive deviation.
Positive Deviation from Raoult’s law. In those non-ideal solutions, when partial pressure of component ‘A’ in the mixture of ‘A’ and ‘B’ is found to be more than that calculated from Raoult’s law. Similarly, the partial vapour pressure of component ‘B’ can be higher than calculated from Raoult’s law.
This type of deviation from ideal behaviour is called positive deviation from Raoult’s law,
e.g., water and ethanol, chloroform and water, ethanol and CCl4, methanol and chloroform, benzene and methanol, acetic acid and toluene, acetone and ethanol, methanol and H2O.
For positive deviation ΔHmixing > 0. (+ ve)
Negative Deviation from Raoult’s law. When the partial vapour pressure of component ‘A’ is found to be less than calculated from Raoult’s law on adding the second component ‘B’. When A is added to B, the partial vapour pressure of solution is less than that of ideal solution of same composition. Boiling point of such a solution is relatively higher than the boiling point of A and B respectively. This type of deviation from ideal behaviour is known as negative deviation from Raoult’s law e.g., chlorofom and acetone, chloroform and methyl acetate, H2O and HCl, H2O and HNO3 acetic acid and pyridine, chloroform and benzene.
For negative deviation ΔHmixing < 0.
Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a specified temperature. It depends
upon the nature of solute and solvent as well as temperature and pressure. In general, a solute dissolves in a solvent if the intermolecular interactions are similar in the two or we may say like dissolves like.
(i) Name the process occurring in the above plant.
(ii) To which container does the net flow of solvent take place?
(iii) Name one SPM which can be used in this plant.
(iv) Give one potential use of the plant.
Answer:
(i) Reverse osmosis.
(ii) Pure water is forced out of the solution to pass through the pores of the membrane in the opposite direction.
(iii) Parchment or cellophone.
(iv) This process is used in desalination to get salt-free water from sea water.
Answer:
Raoult established that the lowering of vapour pressure depends only on the concentration of the solute particles and it is independent of their identity.
establishes a relation between vapour pressure of
the solution, mole fraction and vapour pressure of the solvent, i.e.,
.................1
The reduction in the vapour pressure of solvent (Δp1) is given as:
In a solution containing several non-volatile solutes, the lowering of the vapour pressure depends on the sum of the mole fraction of different solutes.
Equation 3 can be written as :
The expression on the left hand side of the equation as mentioned
earlier is called relative lowering of vapour pressure and is equal to
the mole fraction of the solute. The above equation can be written as:
Here n1 and n2 are the number of moles of solvent and solute respectively present in the solution. For dilute solutions n2 < < n1, hence neglecting n2 in the denominator we have
hence proved
Answer:
According to Raoult's law the partial vapour pressure of each volatile compound in any solution is directly proportional to its mole fraction
if the vapour pressure is higher then solution is said to exhibit positive deviation
Positive deviation is shown by ethanol and water, cyclohexane and ethanol, acetone and diethylether etc.
Reasons:
(i) A—B interaction is weaker than A—A or B—B in positive deviation.
If the vapour pressure is lower then solution then solution is said to exhibit negtive deviation.
Negative deviation is shown by chloroform and acetone, methanol and acetic acid, H2O and HCl, H2O and HNO3 etc.
Reason:
(ii) A—B interaction is stronger than A—A or B—B in negative deviation.
Answer:
Raoult established that the lowering of vapour pressure depends only on the concentration of the solute particles and it is independent of their identity.
establishes a relation between vapour pressure ofthe solution, mole fraction and vapour pressure of the solvent, i.e.
(a) Draw a labelled diagram to show the change in vapour pressure of a solvent, when a non-volatile solute is added to it.
(b) Show that the change in boiling point of the solvent in this diagram?
Answer:
ΔTb = T2 – T1
ΔTb ∝ m
ΔTb = Kb x m
Here m (molality) is the number of moles of solute dissolved in 1 kg of solvent and the constant of proportionality, Kb is called Boiling Point Elevation Constant or Molal Elevation Constant.
Since ΔTb depends on molality of solution and not on nature of solute, therefore, ΔTb is a colligative property.
Answer:
The lowering of vapour pressure of a solution causes a lowering of the freezing point compared to that of the pure solvent.
According to Raoult’s law, when a non-volatile solid is added to the solvent its vapour pressure decreases and now it would become equal to that of solid solvent at lower temperature. Thus, the freezing point of the solvent decreases.
T0f is the freezing point of pure liquid solvent. Tf is the freezing point of solution.
Collegative properties of solution are the properties which depend upon the number of particles present in the solution. When the solute does not undergo any chemical/physical change, the molecular mass of the solute as obtained from the colligative properties is equal to its stoichiometric value.
Therefore in many cases where the solute associate or dissociate in solution certain abnormal values of colligative properties are obtained the value of molecule masses calculated based on colligative properties in such cases will also be abnormal.
Answer:
The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. The ideal solutions have two other important properties. The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is
also zero, i.e.,
ΔmixH = 0, ΔmixV = 0
ideal behaviour of the solutions can be explained by considering two components A and
B. In pure components, the intermolecular attractive interactions will be of types A-A and B-B, whereas in the binary solutions in addition
to these two interactions, A-B type of interactions will also be present.
If the intermolecular attractive forces between the A-A and B-B are nearly equal to those between A-B, this leads to the formation of ideal
solution. example are Solution of n-hexane and n-heptane, bromoethane and chloroethane, benzene and toluene, etc
When a solution does not obey Raoult’s law over the entire range of concentration, then it is called non-ideal solution.
example Mixtures of ethanol and acetone.
Answer:
Vapor pressure or equilibrium vapor pressure is defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system.
(a) When a volatile solute is dissolved in a liquid total vapour pressure is given by the sum of total vapour pressure of both the solvent and solute ie.,
P = PA + PB.
(b) When a non-volatile solute is dissolved in a liquid, the vapour pressure of liquid decreases and this lowering of vapour pressure is proportional to the mole fraction of solute added to it.
osmotic pressure is proportional to the molarity, C of the solution at a given temperature T. Thus:
Π = C R T
Here Π is the osmotic pressure and R is the
gas constant.
Π = (n2 /V) R T
(a) What are non-ideal solution?
(b) What role does the molecular interaction play in deciding the vapour pressure of solutions (i) alcohol and kerosene (ii) Chloroform and acetone.
Answer:
When a solution does not obey Raoult’s law over the entire range of concentration, then it is called non-ideal solution.
(a) Non-ideal solutions are these solutions which
(i) do not follow Raoult’s law.
(ii) ΔHmix ≠ 0.
(iii) ΔVmix ≠ 0.
(iv) The force of attraction between A-A and B-B is not equal to A-B.
(b) In alcohol and acetone, force of attraction is less than alcohol and alcohol molecules as well as acetone and acetone molecule, therefore, vapour pressure increases.
In chlorofom and acetone, force of attraction increases due to intermolecular H-bonding, therefore, vapour pressure decreases.
Answer:
(i) Hemolysis: The red blood cells are protected from the external environment by a semipermeable membrane. The red blood cells are placed in a hypotonic solution. Because the hypotonic solution is less concentrated than the interior of the cell, water moves into the cell. The cell swells and eventually burst, releasing hemoglobin and other molecules. This process is called hemolysis.
(ii) Crenation: When a bacterial cell is placed in a hypertonic (high concentration) sugar solution, the intracellular water tends to move out of the bacterial cell to be more concentrated solution by osmosis. This causes the cell to shrink and eventually, to stop functioning. This process is called crenation.
Answer:
We observe abnormal molecular masses when the solute is an electrolyte and undergo either into association or dissociation.
One unit of an electrolyte compound separates into two or more particles when it dissolves and colligative properties depends on the number of solute particles.
Each NaCl unit dissociates into two ions-Na+ and CI–. Thus, the colligative properties of a 0.1 m solution of NaCl should be twice as great as those of a 0.1 m solution containing a non-electrolyte, such as glucose or sucrose.
Answer:
osmotic pressure is proportional to the molarity, C of the solution at a given temperature T. Thus: Π = C R T
Here Π is the osmotic pressure and R is the gas constant.
Π = (n2 /V) R T
Here V is volume of a solution in litres containing n2 moles of solute. If w2 grams of solute, of molar mass, M2 is present in the solution, then n2 = w2 / M2 and we can write,
in a dilute aqueous solution molarity is equal to molality.
c = m when p = 1 and solution is dilute.
The osmotic pressure will increase with an increase in molality of the solution at a given temperature.
Answer:
The colligative properties of solution depend on the total number of solution.
since the eletrolytes ionise and give more than one particle per formula unit in solution the colligative effect of an eletrolyte solution is alaways greater than that of a non electrolyte of the same molar concentration.
The molar mass of acetic acid in a solution of benzene solvent is greater than its formula molar mass because two CH3COOH molecules form a dimer (CH3COOH)2 in solution. But we know that the molar mass is defined as the mass of 6.023 x 1023 particles of the substance. In case of a normal solute like CH3COOH, the molar mass of acetic acid is equal to the mass of 6.023 x 1023 molecules of formula CH3COOH. While in solution the solution dimerises and becomes a bigger molecule like (CH3COOH)2. Now in this case the molar mass of the solute is equal to the mass of 6.023 x 1023 units of formula (CH3COOH)2. Naturally, this molar mass is larger than the normal formula molar mass. In case of association, Van’t Hoff factor is less than 1, therefore,
MB (abnormal) > MB (normal) when i < 1.
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Answer:
(i) The intermolecular attractive forces between the A-A and B-B are nearly equal to those between A-B.
(ii) A-B interactions are weaker than those between A-A or B-B.
(iii) A-B interactions are more than those between A-A and B-B.
(iv) A-B interactions are less than A-A and B-B interactions.
(v) A-B interactions are more than A-A and B-B interactions.
Answer:
Among H,Cl and F, Hydrogen is least electronegtive while F is most electronegtive than Cl and H.
Thus F can withdraws more electron towards itslef more than Cl and H. So trifluoroacetic acid can easily lose the H+ ions. i.e.
trifluroacetic acid ionize to the larger extent .
Now more the ion produces the greater is the dpress ion of the freezing point
Hence, the depression of freezing point increase in order :
Acetic acid<trichloracetic acid < trifluroacetic acid
Vapour pressures of pure acetone and chloroform at 328 k are 741.8 nm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot Ptotal , Pchlroform and Pacelone as a function of Xactone. The experimental date observed for different composition of mixture is:
|
100 x xacetone |
0 |
11.8 |
23.8 |
36.0 |
50.8 |
58.2 |
64.5 |
72.1 |
|
Pacetone / mm Hg
|
0 |
54.9 |
110.1 |
202.4 |
327.7 |
405.9 |
454.1 |
521.1 |
|
Pchloroform/ mm Hg |
632.8 |
548.1 |
469.4 |
359.7 |
257.7 |
193.6 |
161.2 |
120.7 |
Plot this data also on the same graph paper, indicate whether it has positive deviation or negative deviation from the ideal solution.
It has negative deviation from the ideal solution.Answer:
Raoult’s law states that the partial vapour pressure of a component of a solution at a given temperature is equal to the product of the vapour pressure of the pure component at that temperature and its mole fraction in the solution.
Positive Deviation from Raoult’s law: In those non-ideal solutions, when partial pressure of component ‘A’ in the mixture of ‘A’ and ‘B’ is more than that calculated from Raoult’s law. Similarly, the partial vapour pressure of component ‘B’ can be higher than calculated from Raoult’s law. This type of deviation from ideal behaviour is called positive deviation from Raoult’s law, e.g., water and ethanol, chloroform and water, ethanol and CCl4, methanol and chloroform, benzene and methanol, acetic acid and toluene, acetone and ethanol, methanol and H2O.
For positive deviation ΔHmixing > 0.
Answer:
Henry’s law states that the solubility of a gas in a liquid is directly proportional to pressure of the gas keeping temperature constant.
Mathematically, P = KH x
x mole fraction of gas in solution
P is partial pressure of gas,
KH is Henry's constant.
Applications: Henry's law finds various applications in industry and enables us to explain and understand some biological phenomena. The some of important applications are: CO2 solubility in soft drinks, beverages, soda water etc. is increased by applying high pressure and bottles are sealed under high pressure.
Answer:
Osmotic pressure: Osmotic pressure is the minimum pressure that should be applied to the more concentrated solution to prevent osmosis.
osmotic pressure is proportional to the molarity, C of the solution at a given temperature T.
Thus: Π = C R T
Here Π is the osmotic pressure and R is the gas constant.
Determination of osmotic pressure: Barkley and Hartley’s method: The apparatus consists of a porous pot containing copper ferrocyanide deposited in its wall (acts as semipermeable membrane) and fitted into a bronze cylinder to which is fitted a piston and a pressure gauge (to read the applied pressure).
The pot is fitted with a capillary indicator on left and water reservoir on right. Pot is filled with water while the cylinder is filled with a solution whose osmotic pressure is to be measured. Water tends to pass into the solution through the semipermeable membrane with the result that the water level in the indicator falls down. External pressure is now applied with piston so as to maintain a constant level in the indicator. This external pressure is osmotic pressure.
If the membrane used was a slightly, leaky, then the measured valued of osmotic pressure will not be definite.
Fig. Barkley and Hartley’s apparatus.
Consider 100 g of 10% solution of glucose whose conc. is 10% W / W.
Mass of solution = 100 g
Mass of glucose = 10 g
Mass of solvent = 100 – 10 = 90 g
Molar mass of glucose,
C6H12O6 = 6 x 12 + 12 x 1 + 6 x 16
= 72 + 12 + 96 = 180 g / mol
Molality =
Moles of
Moles of glucose,
Mole fraction of glucose,
Mole fraction of water
Density of sol = 1.2 g mol L
Therefore, volume of solution
Molarity =
Answer:
Let ‘a’ moles of Na2CO3 and a moles of NaHCO3 are present in 1 g equimolar mixture of two. Then we can write (a x 106) + (a x 84) = 1
(Molar masses are : Na2CO3 = 106 and NaHCO3 = 84)
a = 5.26 x 10–3 ‘a’ moles of Na2CO3 = 2
a equivalent of Na2CO3 ‘a’ moles of NaHCO3
=a equivalent of NaHCO3 (2a x 1000) + (a x 1000) = 0.1 x V
∴
[where V is volume of HCl (0.1 M)] Substituting the value of a, we have
1 x 10–4 = 3a = 3 x 5.26 x 10–3
V = 157.8 mL
Mass of solute in 300 g of solution of 25% conc.
Mass of solvent in 300 g of solution
= 300 – 75 = 225 g Moles of solute in 400 g of solution of 40% conc.
Mass of solvent in 400 g of solution
= 400 – 160 = 240 g Total mass of the solute
= 75 + 160 = 235 g Total mass of solvent
= 225 g + 240 g = 465 g Total mass of solution
= 300 + 400 = 700 g Therefore, composition of solute in solution after mixing =
Percentage composition of solvent in solution after mixing
Answer:
Consider ethylene glycol as solute and water as a solvent.
Weight of solute, WB = 222.6 g
Molar mass, MB = 24 + 6 + 32 = 62
WB = 200 g = 0.200 kg.
Moles of solute,
Molaity of ethylene glycol in H2O
Total mass of solution
= 222.6 + 200 = 422.6 g
Density of solution
= 1.072 g/ ml Volume of solution
Molarity of solution,
Answer:
(i) 15 ppm of CHCl3 in water means that there is 15 g of CHCl3 in 106 g of water (1 million = 106)
Percentage of CHCl3
(ii)
Here we have already find the mass % is 15 x 10-4
Number of moles of solute = mass of solute / molar mass
number of moles of CHCl3 =
Molality, m of CHCl3 in drinking water sample
Answer:
We have given,
Molar mass of heptane,
C7H16 = 100 g mol–1
Molar mass of octane,
C8H18 = 114 g mol–1
Moles of heptane
Similarly, Moles of octane
Mole fraction heptane
Mole fraction of octane
Partial vapour pressure = Mole fraction x Vap.
Pressure of pure component.
Partial vapour pressure of heptane
= 0.456 x 105.2 = 47.97 kPa
Partial vapour pressure of octane
= 0.543 x 46.8 = 25.4 kPa
Total vapour pressure of solution = 73.08 kPa.
Answer:
number of moles present in 92g of Na+ ion.
=92/23g mol-1 = 4mol
Molality,
No. of gm moles of solute = 92/23 = 4
Wt. of water (solvent) = 1 kg
Answer:
Mass of solute = 6.5 g
Mass of solution = 450 + 6.5 = 456.6 g
(i) Volume of solution = 1 L = 1000 mL
Density of solution = 1.84 g mL–1
Mass of solution = V x d
= (1000 mL) x 1.84 mL–1
= 18540 g
Mass of H2SO4 = 93g/100mL
93 x 10 = 930 g
Mass of solvent (water)
= 1840 – 930 = 910 g.
(ii) Molarity (m)
Answer:
Mol. mass of benzoic acid, C6H5COOH
= 6 x 12 + 5 x 1 + 12 + 16 + 16 + 1
= 72 + 5 + 12 + 16+16 + 1
= 122 g mol–1
by using formula;
here x= amount of substance required
Amount of benzoic acid required
Answer:
We have given
w2 = 10g
w1 = 250
Kf = 1.86k kg mol-1
M2 = 122.5
Mol. mass of CH3CH2CHCICOOH
Molar Mass of C19 H12 NO3
= 19 x 12 + 21 x 1 + 14 + 48
= 228 + 21 + 14 + 48
= 311 g mol–1
Mole of benzene
Mass of Toluene
Mole fraction of benzene
Heptane and Octane form an ideal solution at 373 K. The vapour pressures of the pure liquids at this temperature are 105.2 K Pa and 46.8 K Pa respectively. If the solution contains 25 g of heptane and 28.5 g of octane, calculate
(i) Vapour pressure exerted by heptane.
(ii) Vapour pressure exerted by solution.
(iii) Mole fraction of octane in the vapour phase.
Components A (Heptane, )
Components of B
No. of moles of heptane,
No. of moles of octane,
Total moles in solution,
(i) Vapour pressure exerted by heptane
where is mole fraction of component A
(ii) Vapour pressure of octane
Total vapour pressure of solution
(iii) Mole fraction of octane
Answer:
Osmotic pressure: Osmotic pressure is the minimum pressure that should be applied to the more concentrated solution to prevent osmosis.
osmotic pressure is proportional to the molarity, C of the solution at a given temperature T. Thus: Π = C R T
Here Π is the osmotic pressure and R is the gas constant.
(a) Measurement of Osmotic Pressure. Different methods are employed for the measurement of osmotic pressure in the laboratory but Berkley and Hartley's method gives the best results. The apparatus consists of a porous pot containing copper ferrocyanide deposited in its wall (acts as semi-permeable membrane) and fitted into a bronze cylinder to which is fitted a piston and a pressure gauge (to read the applied pressure).
The pot is fitted with a capillary indicator on left and water reservoir on right. Pot is filled with water while the cylinder is filled with a solution whose osmotic pressure is to be measured. Water tends to pass into the solution through the semipermeable membrane with the result that the water level in the indicator falls down. External pressure is now applied with piston so as to maintain a constant level in the indicator. This external pressure is osmotic pressure.
If the membrane used was a slightly, leaky, then the measured valued of osmotic pressure will not be definite.
Fig. Berkley and Hartley's apparatus.
(b)
we have given that
mB = 7g
R=0.0821 L atm K-1 mol
Every pure liquid exerts a vapour pressure in the space above it. This is the vapour pressure of the solvent over it at that particular temperature. It depends upon the nature of the solvent and the temperature.
(a) If a volatile solute is dissolved, vapour pressure of the solvent is increased.
(b) However, if a non-volatile solute is dissolved in it, the vapour pressure of the solution is lowered. This is because, in a solution, the percentage of the volatile solvent molecules, which only contributes towards vapour pressure is diminished.
Fig. Decrease of vapour pressure when a non-volatile solute is added to the solvent.
Since, the solute molecules are non-volatile and show no measurable tendency to escape from the solution as vapour, consequently, the vapour pressure of a solution is always lower than that of its solvent.
Raoult’s gave a relation between the relative lowering of vapour pressure and the mole fraction of the solute. Mathematically:
(mole fraction of the solute)
Using the above equation, we can determine the molecular weight of the solute, when the lowering in v.p. is known, when a known weight of the solute w, dissolved in a known wt. of the solvent W.p0 is the vapour pressure of the pure solvent and m and M are the molecular weights of solute and solvent respectively.
Answer:
for any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.
pA ∝ xA
pA x xA where pA is vapour pressure of solvent having mole fraction xA.
PA = P0A x A
But xA + xB = L
∴ xA = 1 – xB
When xB is mole fraction of non-voltile solute B
Pa = P0A (1–xB)
= p0A – p0A x B
Total vapour of solution is equal to pA as nonvolatile solute does not have any vapour pressure.
i.e., Total vapour pressure,
Which of the following solution has the lowest freezing point and why? 0.1m glucose.01 m KCl, 0.1 m Na2SO4.
Answer:
for any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.
pA ∝ xA
Where only solvent is volatile
pA a xA where p A is vapour pressure of solvent having mole fractionxA,
But xA + xB = 1
xA = 1 – xB where xB is mole fraction of non-volatile solute B
pA = p0A (1 – xB)
= p0A – p0A x B
Total vapour pressure
Solution containing non-volatile solute : For a solution of non-volatile solid in a liquid the vapour pressure contribution by the non-volatile solute is negligible. Therefore the partial vapour pressure of a solution containing a non-volatile solute is equal to the product of the vapour pressure of the pure liquid (solvent p0A) and its mole fraction in solution.
PA = P0A x xB ....(i)
xB is the mole fraction of the non-volatile solute
B, then xA + xB = 1
xA = 1 – xB ....(ii)
Substituting the value of xA fromeq. (ii) into eq. (i), we get, pA = p0A (1 – xB) = p0A – p0A x B
Amount of mole of a constituent divided by the total amount of moles of all constituent in a mixture.
Mole fraction of a component =
Henry's law state that
the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution” and is expressed as:
p = KH x
Here KH is the Henry’s law constant
The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. The ideal solutions have two other important properties. The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is also zero, i.e.,
ΔmixH = 0, ΔmixV = 0
van’t Hoff factor modifies the equations for colligative
properties as follows:
Relative lowering of vapour pressure of solvent,
Elevation of Boiling point, ΔTb = i Kb m
Depression of Freezing point, ΔTf = i Kf m
Osmotic pressure of solution, Π = i n2 R T / V
thus it shows that van't hoff factor depend on solute such that it is a colligative properties.
osmotic pressure is proportional to the molarity, C of the solution at a given temperature T. Thus: Π = C R T
Here Π is the osmotic pressure and R is the gas constant.
Π = (n2 /V) R T
Here V is volume of a solution in litres containing n2 moles of solute. If w2 grams of solute, of molar mass, M2 is present in the solution, then n2 = w2 / M2 and we can write,
in a dilute aqueous solution molarity is equal to molality.
c = m when p = 1 and solution is dilute.
The osmotic pressure will increase with an increase in molality of the solution at a given temperature.
Van’t Hoff introduced a factor ‘i’ called Van’t Hoff’s factor, to express the extent of association or dissociation of solutes in solution. It is ratio of the normal and observed molecular masses of the solute, i.e
In case of association, observed molecular mass being more than the normal, the factor i has a value less than 1. But in case of dissociation, the Van’t Hoff’s factor is more than 1 because the observed molecular mass has a lesser value than the normal molecular mass. In case there is no dissociation the value of ‘i’ becomes equal to one.
Answer:
(i) Liquid in liquid: Ethanol dissolved in water
(II) Gas in gas : Mixture of oxygen and nitrogen gases
Answer:
osmotic pressure is proportional to the molarity, C of the solution at a given temperature T. thus
Π = C R T
Here Π is the osmotic pressure and R is the
gas constant.
Π = (n2 /V) R T
Here V is volume of a solution in litres containing n2 moles of solute.
If w2 grams of solute, of molar mass, M2 is present in the solution, then
n2 = w2 / M2 and we can write
Thus, knowing the quantities w2, T, Π and V we can calculate the molar mass of the solute.
Answer:
Azeotropes are a mixture of at least two different liquids. Their mixturecan either have a higher boiling point than either of the components or they can have a lower boiling point. Azeotropes occur when fraction of the liquids cannot be altered by distillation.
example ethanol (95.5%) + water (4.5) mixture boils at 351.5k
Answer:
Relative lowering vapour pressure and molality are related by the equaton:
where,
m = molality of the solution
MA = Molar mass of the solvent.
Answer:
Osmotic pressure is the minimum pressurewhich needs to be applied to a solution to prevent the inward flow of water across a semipermeable membrane. It is also definedas the measure of the tendency of a solution to take in water by osmosis.
or
osmotic pressure is proportional to the molarity, C of the solution at a given temperature T. Thus:
Π = C R T
Here Π is the osmotic pressure and R is the
gas constant.
Answer:
Raoult's law states that for a solution of volatile liquid, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.
Thus if there is solution of two liquid component 1and 2, then
for component 1,
Where
p1p2 = partial vapour pressure of two volatile component in solution 1 and 2
= vapour pressure of pure component1 and 2
x1x2 = mole fraction of component.
Answer:
Lower A has lower boiling point so it is being more volatile and has high vapour pressure at 800 C. Becausevapour pressure has an inverse relation with boiling point.
Answer:
In the same volume of water,0.1 molar of NaCl will have a greater boiling point elevation, as boiling point elevation is a colligative property which depends on the relative amount of the constituent and not on their identity.
Since more ions will be produced by NaCl (ionic compound) than glucose (covalent compound)
as there are more componets part in the NaCl molecule so the NaCl will have more boiling point.
Answer:
It is because ethyl alchol and water do not form an ideal solution the intermolecular forces of attraction are weak so the molecules will become far apart from each other hence volume wll increase.
Answer:
(i) The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. The ideal solutions have two other important properties. The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is also zero, i.e.,
ΔmixH = 0
ΔmixV = 0
It means that no heat is absorbed or evolved when the components are mixed. Also, the volume of solution would be equal to the sum of
volumes of the two components.
(ii)
At molecular level, ideal behaviour of
the solutions can be explained by considering two components A and B. In pure components, the intermolecular attractive interactions will
be of types A-A and B-B, whereas in the binary solutions in addition to these two interactions, A-B type of interactions will also be present.
If the intermolecular attractive forces between the A-A and B-B are nearly equal to those between A-B, this leads to the formation of ideal
solution.
example: Solution of n-hexane and n-heptane, bromoethane and chloroethane, benzene and toluene, etc.
Escaping tendency of 'A' and 'B' should be same in pure liquids and in the solution.
Answer:
when a non-volatile solid is added to the solvent
its vapour pressure decreases and now it would
become equal to that of solid solvent at lower
temperature. Thus, the freezing point of the
solvent decreases
Answer:
Positive deviation from a Raoult's law:
mixture of ethanol and cyclohexane show positive deviation.
when pure ethanol is added to cyclohexane the moleculeof cyclohexane come in between the molecule of ethanol which result in the breaking of hydrogen bond and thus positive deviation.
Negative deviation from Raoult's law:
Mixture of chloroform and acetone shows negtive deviation when these are mixed hydrogen bond is formed and negtive deviation is shown:
Answer:
Raoult's law states that for a solution of volatile liquid, the partial vapour pressure of each component in the solution is dirctly proportional to its mole fraction.
Thus if there is solution of two liquid component 1 and 2 then,
for component 1,
Answer:
In case of negative deviations from Raoult’s law, the intermolecular attractive forces between A-A and B-B are weaker than those between A-B and leads to decrease in vapour pressure resulting in negative
deviations. An example of this type is a mixture of phenol and aniline.
In this case the intermolecular hydrogen bonding between phenolic proton and lone pair on nitrogen atom of aniline is stronger than the respective intermolecular hydrogen bonding between similar
molecules.
This decreases the escaping tendency of molecules for each component and consequently the vapour pressure decreases resulting
in negative deviation from Raoult’s law
Answer:
A molar mass that is either lower or higher than the expected or normal value is called as abnormal molar mass.
In order to account for all such abnormalities, introduced a factor
(i) known as van 't Hoff's factor, which represents the extent of association (or) dissociation of a solute.
Van't Hoff's factor is defined as the ratio of the observed colligative property to the calculated colligative property.
i = Observed colligative property / Calculated colligative property
Observed colligative property ∝ 1/Molar Mass
i = Mc/Mo
Van't Hoff's factor
(i) represents the extent of association (or) dissociation of a solute
i = Total number of moles of particles after association or dissociation / Number of moles of particles before association or disscussion
Experimentally determined molar mass is always lower than actual value for dissociation.
Molar Mass ∝ 1/Colligative Property
If the solute undergoes association in a solution, then the value of van 't Hoff's factor is less than one. If the solute undergoes dissociation then 'i' is greater than one.
Ex:
KCl → K+ + Cl-
1 molecule of KCl furnishes 2 ions in solution
i = Total number of moles of particles after dissociation / Number of moles of particles before dissociation
i = 2/1 = 2
2CH3COOH ⇔(CH3COOH)2
Ethanoic acid Dimer of Ethanoic acid
i = Total number of moles of particles after association / Number of moles of particles before association
i = 1/2 = 0.5
Answer:
The solution of ethanol and cyclohexane show positive deviation.
this is because when pure ethanol is added to cyclohexane the molecule of cyclohexane come in between the molecule of ethanol which result in the breaking f hydrogen bond and thus positive deviation .
another case can be
the mixture of chloroform and acetone shows negtive deviation.
When these are mixed , hydrogen bond is formed and negtive deviation shown;
Answer:
When a solution does not obey Raoult’s law over the entire range of concentration, then it is called non-ideal solution.
The vapour pressure of such solution either be higher or lower. i.e positive when higher
negtive when lower.
The cause for these deviations lie in the nature of interactions at the molecular level.
In case of positive deviation from Raoult’s law,
A-B interactions are weaker than those between A-A or B-B, i.e., in this case the intermolecular attractive forces between the solute-solvent molecules are weaker than those between the solute-solute and solvent-solvent molecules. This means that in such solutions, molecules of A (or B) will find it easier to escape than in pure state. This will increase the vapour pressure and result in positive deviation. Mixtures of ethanol and acetone behave in this manner.
In case of negative deviations from Raoult’s law, the intermolecular attractive forces between A-A and B-B are weaker than those between
A-B and leads to decrease in vapour pressure resulting in negative deviations. An example of this type is a mixture of phenol and aniline.
In this case the intermolecular hydrogen bonding between phenolic proton and lone pair on nitrogen atom of aniline is stronger than the
respective intermolecular hydrogen bonding between similar molecules.
Answer:
Liquids at a given temperature vapourise and under
equilibrium conditions the pressure exerted by the vapours of the liquid over the liquid phase is called vapour pressure.
When a volatile solute is dissolved into, solvent then the vapour state solute and the solvent.
total vapour pressure above such a solution will be equal to the sum of the pressure exerted by the vapours of both solute and solvent.
Where
Ptotal = total pressure of solution
(b)
When a non- volatile solute is dissovled then there is lowering in vapour pressure .
The lowering in vapour pressureisgiven by
Answer:
Elevation of boiling point with addition of non-volatile solute vapour pressure decrese and hence boiling point increase.
Let be the boiling point of pure solvent and
Tb be the boiling point of solution. The increase in
the boiling point is known as
elevation of boiling point.
for dilute
solutions the elevation of boiling point (ΔTb) is
directly proportional to the molal concentration of
the solute in a solution. Thus
ΔTb ∝ m
or
ΔTb = Kb m
Here m (molality) is the number of moles of solute dissolved in 1 kg of solvent and the constant of proportionality, Kb is called Boiling Point Elevation Constant or Molal Elevation Constant. The unit of Kb is K kg mol-1.If w2 gram of solute of molar mass M2 is dissolved in w1 gram of solvent, then molality, m of the solution is given by the expression:
Thus, in order to determine M2, molar mass of the solute, known mass of solute in a known mass of the solvent is taken and ΔTb is
determined experimentally for a known solvent whose Kb value is known.
Answer:
Let us calculate the concentration of the first solution with osmatic pressure 4.98bar
mass of glucose = 36g
molar mass of gulcose = 180g/ mol
therefore number of moles of gulcose = 36/180
= 0.2 moles
volume of the solution = 1L
molarity = No. of moles of glucose / vol.of solution
molarity =0.2/1L
C1 = 0.2 moles/L
Answer:
The molality of the cane sugar, m= 0.1539m
depression in freezing point = 273.15-271
=2.15K
Since = Kfm
or Kf = /m = 2.15k/0.1539m = 13.97K/m
Now the weight of glucose , W2 = 5g
molecular mass of glucose, M2 =180g/mol
then =Kfm
=
then freezing point of solution = 273.15-3.88
=263.27K
Answer:
The vapour pressure of a pure solvent decrease .
when a non-volatile solute is added to the solvent
this is because on adding the solute a fewer number of water molecules are present at the surface which can evaporate as some of the area is occupied by -non- volatile solute molecules thereby decreasing the vapour pressure of the solution of the glucose in wateris lower than pure water.
Answer;
Let the density of solution = d g/cm3
and volume of solution is = 1L = 1000 cm3
mass of the solution = d xV
= (1000d)g
6.90 M solution mean 1L solution contains 6.90 moles of KOH
therefore mass of KOH = 6.90 x 56 = 386.4g
but only 30% of the solutionby mass is KOH
therefore
30/100 x (1000)d =386.4
and density is 1.288g/ cm3
Answer:
When 30 ml of ethyl alcohol and 30 ml of water are mixed, the volume of resulting solution is more than 60ml.
It is because ethyl alchol and water do not form an ideal solution intermolecular forces of attraction are weak so the molecule will become farapart from each other hence volume will increase.
Answer:
Copper sulphate conduct electricity only in molten state or in aqueous phase because on dissolving or melting it. copper sulphate dissociate in copper ion and sulphate ion and became free to move.
As they are freeto move in molten and aqueous phase the charge particle conduct electricity.
In solid and any other ionic form the ions held in a lattice and unable to move so they are not conduct electricity.
Answer:
by using formula .
let the mass of solution be 100g
therefore mass of urea = 10g
molecular mass of urea = 60 g
xurea = 10/60 = 1/6
molecular mass of water = 18 g
xwater = 90/18 = 5
mole fraction of urea =
Answer:
0.1M NaCl and0.1M glucose solution means that 0.1 moles of solute dissolve in 1L of solvent.
weight of solvent in both solution is same .
Dpression in freezing point can br calculated as :
Where
Kf = molal depression constant
wB = weight of solute
wA= weight of solvent
MB = molar mass of the solute
mass of NaCl = 58.69g/mol
mass of glucose = 180g/mol
The depression in freezingpoint is inversly proprotional to molar mass of solute thus more is molar mass of solute lesser is the depression in freezing point .
Thus it can explain that the NaCl wil be more than the glucose.
Answer;
Let the mass of solution =100g
xKOH = 30/56
volumeof solution =
Answer:
= xurea
Let the mass of solution be 100g
therefore mass of urea = 10 g
molecular mass of urea = 60 g
xurea = 10/60 =1/6
molecular mass of water = 18g
xwater = 90/18 =5
The relative lowering of vapour pressure is given by the following expression.
Where is the vapour pressure of pure solvent, Psolution is the vapour pressure of solution containing dissolved solute, n1 is the number of moles of solvent and n2 is the number of moles of solute.
For dilute solution n2<<n1, therefore the above expression reduces to
Where w1 and w2 are the masses and M1 and M2 are the molar masses of solvent and solute respectively.
We are given that
W1 =30 g
W2 =30g
psolution =2.8 kpa
p0solvent =? And M2=?
Answer:
Calculate the vapour- pressure lowering of water when5.67g of glucose, C6H12O6 is dissolved in 25.2 g water at 250C. The vapour pressure of water at 250C is 23.8 mmHg. What is the vapour pressure of the solution?
WB =5.67g
WA =25.2g
MB=180g/mol
MA=18g/mol
NB= 5.67/180 =0.0315 mol glucose
NA=25.2/18 =1.40 mol water
Answer:
AS both the solution are isotonic they should have same concentrationin mole/litre
for sucrose solution concentration =4g/100cm3
= 40g/Litre
molar mass of sucrose C 12H22O11 = 342
therefore we get =40/342 moles/litre
For unknown substance Let N be the molecular mass then concentration = 3g/100cm3
= 30g/Litre
= 30/N moles/Litre
comparing the both equation
30/N = 40/342
N=(30 x 342)/40
N= 256.5
Molecular mass of oragnic compound
Answer:
13 % of solution of sulphuric acid contains 13 g of H2SO4in 100 g of the solution. Weight of water = 100 - 13 = 87 g.
mole of H2SO4 = 13/98 mole
Volume of H2SO4 solution + Weight of soluton / Density = 100/1.02 ml
= 100 / 1.02 x 1/1000 = 1/1.02 litre
Molarity of H2SO4 solution = mole of H2SO4 / Volume of solution
= 13/98 / 1/1.02 = 10.2 x 13/98 = 1.353 M
Molality of H2SO4 solution = mole of H2SO4 / wt of water in kg. = 13/98 / 87/1000 = 13/98 x 1000/87
= 1.525 mol kg-1 (mole Kg-1)
Answer:
osmotic pressure is proportional to the molarity, C of the solution at a given temperature T. Thus:
Π = C R T
Here Π is the osmotic pressure and R is the
gas constant.
Π = (n2 /V) R T
Here V is volume of a solution in litres containing n2 moles of solute.
If w2 grams of solute, of molar mass, M2 is present in the solution, then
n2 = w2 / M2 and we can write,
Thus, knowing the quantities w2, T, Π and V we can calculate the molar mass of the solute.
Answer:
In case of positive deviation from Raoult’s law, A-B
interactions are weaker than those between A-A or B-B, i.e., in this case the intermolecular attractive forces between the solute-solvent molecules
are weaker than those between the solute-solute and solvent-solvent molecules. This means that in such solutions, molecules of A (or B) will
find it easier to escape than in pure state. This will increase the vapour pressure and result in positive deviation. Mixtures of ethanol and acetone is good example of this.
Answer:
the vapour pressure of a liquid increases with increase of temperature. It boils at the temperature at which its vapour pressure is equal to the atmospheric pressure. For example, water boils at 373.15 K (100° C) because at this temperature the vapour pressure of water is 1.013 bar (1 atmosphere)
Let be the boiling point of pure solvent and
Tb be the boiling point of solution. The increase in
the boiling point is known as
elevation of boiling point.
for dilute solutions the elevation of boiling point (ΔTb) is directly proportional to the molal concentration of the solute in a solution. Thus
ΔTb ∝ m
or ΔTb = Kb m
Here m (molality) is the number of moles of solute dissolved in 1 kg of solvent and the constant of proportionality, Kb is called Boiling Point Elevation Constant or Molal Elevation Constant (EbullioscopicConstant). The unit of Kb is K kg mol-1. If w2 gram of solute of molar mass M2 is dissolved in w1 gram of solvent, then molality, m of the solution is given by the expression:
Thus, in order to determine M2, molar mass of the solute, known mass of solute in a known mass of the solvent is taken and ΔTb is
determined experimentally for a known solvent whose Kb value is known.
Answer:
the freezing point of a substance may be defined as the temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase.
According to Raoult’s law, when a non-volatile solid is added to the solvent its vapour pressure decreases and now it would become equal to that of solid solvent at lower temperature. Thus, the freezing point of the solvent decreases.
Let be the freezing point of pure solvent and Tf be its freezing point when non-volatile solute is dissolved in it. The decrease in freezing point.
Similar to elevation of boiling point, depression of freezing point (ΔTf) for dilute solution (ideal solution) is directly proportional to molality,
m of the solution. Thus,
ΔTf ∝ m
or
ΔTf= Kfm
The proportionality constant, Kf, which depends on the nature of the solvent is known as Freezing Point Depression Constant or Molal Depression Constant or Cryoscopic Constant. The unit of Kf is K kg mol-1
If w2 gram of the solute having molar mass as M2, present in w1 gram of solvent, produces the depression in freezing point ΔTf of the
solvent then molality of the solute is given by the equation:
Thus for determining the molar mass of the solute we should know the quantities w1, w2, ΔTf, along with the molal freezing point depression constant.
Answer:
Raoult's law state that for any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.
p1 ∝ x1
and p1 = x1
The proportionality constant is equal to the vapour pressure of pure solvent, .
IDEAL SOLUTION
(i) The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. The ideal solutions have two other important properties. The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is also zero, i.e.,
ΔmixH = 0
ΔmixV = 0
It means that no heat is absorbed or evolved when the components are mixed. Also, the volume of solution would be equal to the sum of
volumes of the two components.
(ii)
At molecular level, ideal behaviour of
the solutions can be explained by considering two components A and B. In pure components, the intermolecular attractive interactions will
be of types A-A and B-B, whereas in the binary solutions in addition to these two interactions, A-B type of interactions will also be present.
If the intermolecular attractive forces between the A-A and B-B are nearly equal to those between A-B, this leads to the formation of ideal
solution.
example: Solution of n-hexane and n-heptane, bromoethane and chloroethane, benzene and toluene, etc.
NON-IDEAL SOLUTION
When a solution does not obey Raoult’s law over the entire range of concentration, then it is called non-ideal solution.
In non-ideal solution the vapour pressure of such a solution is either higher or lower than that predicted by Raoult’s law, If it is higher, the solution exhibits positive deviation and if it is lower, it exhibits negative deviation.
IN case of non-ideal solution
ΔmixH ,ΔmixV
both are not equal to zero. it either be less than zero or greater than zero.
IN case of positive deviation it be larger and In case of negtive deviation it be lesser.
in the binary solutions in addition to these two interactions, A-B type of interactions will also be present.
If the intermolecular attractive forces between the A-A and B-B are not nearly equal to those between A-B, this leads to the formation of non ideal solution.
example :a mixture of chloroform and acetone etc.
What is meant by positive deviations from Raoult's law? Give an example. What is the sign of mixH for positive deviation?
OR
Define azeotropes. What type of azeotrope is formed by positive deviation from Raoult's law? Give an example.
The solutions that do not obey Raoult’s law over the entire range of concentration are known as non-ideal solutions. They have vapour pressures either higher or lower than those predicted by Raoult’s law. If the vapour pressure is higher, then the solution is said to exhibit a positive deviation from Raoult’s law.
A mixture of ethanol and acetone is an example of a solution that shows a positive deviation from Raoult’s law.
In case of solutions showing positive deviations, absorption of heat takes place that mixH has a positive sign.
OR
Azeotropes are the binary mixtures of solutions that have the same composition in liquid and vapour phases and that have constant boiling points.
A minimum-boiling azeotrope is formed by solutions showing a large positive deviation from Raoult’s law at a specific composition.
Example: An ethanol–water mixture containing approximately 95% ethanol by volume
3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the van't Hoff factor and predict the nature of solute (associated or dissociated).
(Given: Molar mass of benzoic acid = 122 g mol-1, Kf for benzene = 4.9 K kg mol-1
We know that the depression in freezing point is given by
Here
Depression in freezing point K
Kf for benzene=4.9kg mol-1
Mass of benzene W=49g
Molar mass of benzoic acid ms = 122g mol-1
Substituting the value we get
1.62 = 
As the value of i<1 benzoic acid is associated solute.
Calculate the mass of compound (molar mass = 256 g mol-1) to be dissolved in 75 of benzene to lower its freezing point by 0.48 K (Kf= 5.12 kg mol-1 )
We have given:
Tf = 0.48 K
Kf=5.12K kg mol-1
W1 = 75 g
W2 = ?
M2 = 256 g mol-1
Define an ideal solution and write one of its characteristics.
Definition: The solutions that obey Raoult’ s law over the entire range of concentration are called ideal solutions. Examples: n-hexane and n-heptane.
Characteristics: For ideal solutions:
Enthalpy of mixing (mixH) of the pure components to form the solution is zero.
The volume of mixing (mixV) is also equal to zero.
18 g of glucose, C6H12O6 (Molar Mass = 180 g mol-1) is dissolved in 1 kg of water in a saucepan. At what temperature will this solution boil?
(Kb for water = 0.52 K kg mol-1, boiling point of pure water = 373.15 K)
w1 = weight of solvent (H2O) = 1 kg
w2 = weight of solute (C6H12O6) = 18 gm
M2 = Molar mass of solute (C6 H12O6) = 180 g mol-1
Kb = 0.52 K Kg mol-1
Determine the osmotic pressure of a solution prepared by dissolving 2.5 x 10-2 g of K2SO4 in 2L of water at 250C, assuming that it is completely dissociated.
(R = 0.0821 L atm K-1 mol-1, Molar mass of K2SO4 = 174 g mol-1)
w2 = 2.5 x 10-2 g (Mass of K2SO4)
M2 = 174 g mol-1 (Molar mass of K2SO4)
V = 2L,
R = 0.0821 L atm K-1 mol-1 and
T = 25°C = 298 K
Osmotic pressure,
A 1.00 molar aqueous solution of trichloroacetic acid (CCl3COOH) is heated to its boiling point. The solution has the boiling point of 100.180C. Determine the van’t Hoff factor for trichloroacetic acid. (Kb for water = 0.512 kg mol-1)
Molality of solution = m = 1.00 m
Boiling points of solution = Tb = 100.180C = 373.18 K
Boiling point of water (solvent) = 
= 100.00° C = 373 K
Elevation in boiling point = 
- Tb
Observed boiling point = 373.18 K - 373 K = 0.18 K
Kb water = 0.512 K kg mol- 1
∴ 
T b= Kb x m
= 0.512 x 1 = 0.512 K
∴ Calculated boiling point = 0.512 K
Define the following terms:
(i) Mole fraction
(ii) Isotonic solutions
(iii) Van’t Hoff factor
(iv) Ideal solution
(i) Mole fraction: The mole fraction of a component in a mixture is defined as the ratio of the number of moles of the component to the total number of moles of all the components in the mixture. Mathematically, it is represented as:
Mole fraction is denoted by ‘x’.
(ii) Isotonic solution: It is a type of solution that has the same salt concentration as its surrounding environment and thus the substances around it neither lose nor gain water by osmosis.
(iii) Van’t Hoff factor: It is defined as the ratio of the experimental value of colligative property to the calculated value of the colligative property and is used to find out the extent of dissociation or association.
Mathematically, it is represented as:
For association, i < 1
For dissociation, i > 1
No association or dissociation, i = 1
Examples: One formula unit of NaCl will create two particles in solution, a Na+ ion and a Cl- ion.
One formula unit of CaCl2 will create three particles in solution, a Ca+ ion and two Cl- ions.
(iv) Ideal Solutions: Ideal Solutions are those which obey Raoult's Law at all concentrations and Temperatures. Some examples of ideal solution liquid pairs are benzene and toluene, n-heptane and n-hexane, ethyl bromide and ethyl iodide, chlorobenzene and bromobenzene etc.
(a) Differentiate between molarity and molality for a solution. How does a change in temperature influence their values?
(b) Calculate the freezing point of an aqueous solution containing 10.50 g of MgBr2in 200 g of water. (Molar mass of MgBr2 = 184 g) (Kf for water = 1.86 K kg mol-1)
Molarity is defined as the number of moles of solute dissolved per liter of solution.
Molality of a solution is defined as the number of moles of solute dissolved in 1000 grams of solvent.
While molarity decreases with an increase in temperature, molality is independent of temperature. This happens because molality involves mass, which does not change with a change in temperature, while molarity involves volume, which is temperature dependent.
b) Given
w2 = 10.50 g
w1 = 200 g
Molar mass of MgBr2 (M2) = 184 g
Using the formula,
(a) Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a colligative property? Explain.
(b) Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0 g of water. (Kb for water = 0.512 K kg mol-1), (Molar mass of NaCl = 58.44 g)
(a) Osmosis: The process of flow of solvent molecules from pure solvent to a solution or from a solution of lower concentration to a solution of higher concentration through a semi-permeable membrane is called osmosis.
Osmotic Pressure: The pressure required to just stop the flow of solvent due to osmosis is called osmotic pressure (π) of the solution.
Yes, the osmotic pressure of a solution is a colligative property. The osmotic pressure is expressed as.
Where,
π = osmotic pressure
n = number of moles of solute
V = volume of solution
T = temperature
From the equation, it is clear that osmotic pressure depends on upon the number of moles of solute ‘n’ irrespective of the nature of the solute.
Hence, osmotic pressure is a colligative property.
(b) Given
Kb = 0.512 k kg mol-1
w2 = 15.00 g
w1 = 250.0 g
M2 = 58.44 g
Using the formula,
Why does a solution containing non-volatile solute have higher boiling point than the pure solvent? Why is elevation of boiling point a colligative property?
The boiling point of substance can be defined as, it is the temperature at which the vapour pressure of the substance equal to atmospheric pressure. According to Raoult's law, the vapour pressure of a solvent decreases in the presence of a non-volatile solute. Thus, the vapour pressure of a solution containing a non-volatile solute requires a high temperature to become equal to the atmospheric pressure. That is why the boiling point of a solution containing a non-volatile solute increases. Thus, the solution containing a non-volatile solute has a higher boiling point than a pure solvent. Example of non-volatile solute sugar, NaCl. etc.
Elevation in boiling point is a colligative property, as it depends on the number of solute particles present in a solution.
Calculate the freezing point of the solution when 31 g of ethylene glycol (C2H6O2) is dissolved in 500 g of water. (Kf for water = 1.86 K kg mol–1)
The freezing point of pure water is 273.15 K. On dissolving ethylene glycol, freezing point, being a colligative property, will be lowered.
We are given that
Kf for water =1.86 K kg mol-1
Mass of solute, ws = 37 g
Molar mass of solute, Ms = 12 x 2 + 1 x 6 + 16 x 2
= 62 g mol-1
Mass of water, W = 500 g
Therefore 
Hence, the freezing point of the solution, Tf = 273.15K – 1.86 K
= 271.29 K
(a) State Raoult’s law for a solution containing volatile components.
How does Raoult’s law become a special case of Henry’s law?
(b) 1·00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0·40 K. Find the molar mass of the solute. (Kf for benzene = 5·12 K kg mol-1)
(a) Let p1, p2 = Partial vapour pressure of two volatile components 1 and 2 of a mixture 
= Vapour pressure of pure components 1 and 2
x1, x2 = Mole fractions of the components 1 and 2
ptotal = Total vapour pressure of the mixture then Raoult’s law can be stated as: For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.
That is, for component 1,
p1 
x1
And, p1 = 
x1
For component 2,
P2 = 
x2
According to Dalton’s law of partial pressures,
The plot of vapour pressure and mole fraction of an ideal solution at constant temperature is shown below.
Raoult’s Law as a Special Case of Henry’s Law
According to Raoult’s law, the vapour pressure of a volatile component in a given solution is p1 = 
x1
According to Henry’s law, the partial vapour pressure of a gas (the component is so volatile that it exists as gas) in a liquid is
p = KH x
It can be observed that in both the equations, the partial vapour pressure of the volatile component varies directly with its mole fraction. Only the proportionality constants KH and are different. Thus, Raoult’s law becomes a special case of Henry’s law in which KH is equal to .
(b) Given: w2 = 1g (weight of solute)
w1 = 50 g (weight of solvent)
Tf = 0.40 K
kf = 5.12 K Kg mol-1
M2 =? (Molar mass of solute)
Using the equation,
Tf = Kfm (where m is molality)
0.40 = 5.12 x m
(a) Define the following terms:
(i) Ideal solution
(ii) Azeotrope
(iii) Osmotic pressure
(b) A solution of glucose (C6H12O6) in water is labeled as 10% by weight. What would be the molality of the solution?
(Molar mass of glucose = 180 g mol-1)
(a)
(i) Ideal Solution:
Solutions which obey Raoult’s law over the entire range of concentrations are known as the ideal solution. Along with that for ideal solution:
Enthalpy of mixing of the pure components to form the solution i.e 
mix H = 0 and volume of mixing, 
mix V = 0.
An ideal solution will be formed when intermolecular forces of attraction between the molecules of solute (A - A) and those between the molecules of solvent (B -B) are nearly equal to those between solute and solvent molecules (A - B).
For Example n-Hexane and n-heptane
(ii) Azeotropes
Binary mixtures which have the same composition in liquid and vapour phase, and have constant boiling points are known as azeotropes. It is not possible to separate its components by fractional distillation .There are two types of azeotropes:
Minimum boiling azeotrope, example: Ethanol-water mixture containing ethanol approximately 95% by volume.
Maximum boiling azeotrope, example: Nitric acid-water mixture containing 68% nitric acid and 32% water by mass.
(iii) Osmotic Pressure
The process of flow of solvent molecules from pure solvent to a solution or from a solution of lower concentration to a solution of higher concentration through a semi-permeable membrane is called osmosis. The pressure required to prevent the flow of solvent due to osmosis is called osmotic pressure (Ï€) of the solution.
Osmotic pressure is directly proportional to the molarity C of the solution at a given temperature T.
Where,
n2 = Number of moles of solute
V= Volume of the solution in litres
(b) Assume that 100 g of solution contains 10 g of glucose and 90 g of water as our glucose solution is 10% by weight
Where w2 = weight is solute in 'g'
w1 = weight of solvent in 'g'
M2 = Molecular mass of solute.
So,
a) Define the following terms:
(i) Mole fraction
(ii) Ideal solution
(b) 15.0 g of an unknown molecular material is dissolved in 450 g of water. The resulting solution freezes at - 0.34°C. What is the molar mass of the material? (Kf for water = 1.86 K kg mol-1)
(i) Mole fraction: The mole fraction of a component in a mixture is defined as the ratio of the number of moles of the component to the total number of moles of all the components in the mixture. Mathematically, it is represented as:
Mole fraction is denoted by ‘x’.
(ii) Ideal Solution:
Solutions which obey Raoult’s law over the entire range of concentrations are known as ideal solution. Along with that for ideal solution:
Enthalpy of mixing of the pure components to form the solution i.e mix H = 0 and volume of mixing, mix V = 0.
An ideal solution will be formed when intermolecular forces of attraction between the molecules of solute (A - A) and those between the molecules of solvent (B -B) are nearly equal to those between solute and solvent molecules (A - B).
For Example: n-Hexane and n-heptane
(b) Given mass of solute = wb = 15.0g
Molar mass of solute = Mb =?
Mass of water = wa = 450 g
Freezing point of water = 0°C = 273 K
Freezing point of solution = - 0.34°K
= (- 0.34 + 273) K
Depression in freezing point = Tf= 273 - (-0.34 + 273)
= 0.34 K
Kf for water = 1.86 K Kg mol-1
(a) Explain the following:
(i) Henry’s law about the dissolution of a gas in a liquid.
(ii) Boiling point elevation constant for a solvent.
(b) A solution of glycerol (C3H8O3) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42°C. What mass if glycerol was dissolved to make this solution? (Kb for water = 0.512 K kg mol-1)
(a)
(i) Henry’s law states that partial pressure of a gas in the vapour phase is proportional to the mole fraction of the gas in the solution. If p is the partial pressure of the gas in the vapour phase and x is the mole fraction of the gas, then Henry’s law can be expressed as:
p = KHx
Where,
KH is Henry’s law constant
(ii) The boiling point elevation constant or molal elevation constant is a constant quantity for a solute which is related to molar mass and elevation in boiling point by the following relation.
Where, Kb is the boiling point elevation constant
MB is the molar mass of the solute
WB is the weight of the solute
WA is the weight of the solvent
Tb is the elevation in boiling point
(b) WB =?
WA = 500g
Kb = 0.512 Kkg mol-1
Tb = 100.42°C - 100°C
= 0.42°C
MB = 3 x 12 + 8 x 1 + 3 x 16
= 36 + 8 + 48 = 92
a) Calculate the freezing point of the solution when 1.9 g of MgCl2 (M = 95 g mol−1) was dissolved in 50 g of water, assuming MgCl2 undergoes complete ionization.
(Kf for water = 1.86 K kg mol−1)
(b)
(i) Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why?
(ii) What happens when the external pressure applied becomes more than the osmotic pressure of solution?
OR
(a)When 2.56 g of sulphur was dissolved in 100 g of CS2, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (Sx).
(Kf for CS2 = 3.83 K kg mol−1, Atomic mass of sulphur = 32 g mol−1]
(b)Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing
(i)1.2% sodium chloride solution?
(ii)0.4% sodium chloride solution?
(i) The elevation in the boiling point of a solution is a colligative property; therefore, it is affected by the number of particles of the solute. Since the amount of solute is higher in 2 M glucose solution as compared to 1 M glucose solution, the elevation in the boiling point is higher. Hence, 2 M glucose solution has a higher boiling point than 1 M glucose solution.
(ii) When the external pressure exerted on the solution is higher than the osmotic pressure, the pure solvent starts flowing out of the solution through the semi- permeable membrane. This process is known as reverse-osmosis.
Or
(b)
(i) 1.2% sodium chloride solution is hypertonic wrt 0.9% sodium chloride solution. Therefore, when the blood cells are placed in 1.2% sodium chloride solution, water flows out of the cells and the cells shrink.
(ii) 0.4% sodium chloride solution is hypotonic wrt 0.9% sodium chloride solution. Therefore, when the blood cells are placed in 0.4% sodium chloride solution, water flows into the cells and the cells swell.
(i) Gas (A) is more soluble in water than Gas (B) at the same temperature which one of two gases will the higher value of KH (Henry's constant) and why ?
(ii)In non-ideal solution, what type of deviation shows the formation of maximum boiling azeotropes?
(i)According to Henry's law , at the constant temperature, the solubility of the gas in a liquid is directly proportional to the partial pressure of the gas on the surface of the liquid or solution. Gas B has a Higher value of KH and thus lower is the solubility of the gas.
p = KHx
(ii) Negative deviation from Raoult’s law
What is meant by ‘reverse osmosis’?
The phenomenon of reversal of the direction of osmosis by the application of a pressure larger than the osmotic pressure on the solution side is known as reverse osmosis. In this case, the pure solvent flows out of the solution through a semi-permeable membrane.
Differentiate between molarity and molality values for a solution. What is the effect of change in temperature on molarity and molality values?
Molarity is defined as the number of moles of solute dissolved per litre of solution.
Molality of a solution is defined as the number of moles of solute dissolved in 1000 grams of solvent.
A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr at 25°C.
Assuming that the gene fragment is a non-electrolyte, calculate its molar mass.
The molar concentration of the gene fragment
Here, WB = 8.95 mg = 8.95 x 10-3 g, R = 0.0821 L atm Mol-1 K-1
T = 250C = (25 + 273) K = 298 K. = 0.355torr =0.355/760 atm
V = 35 mL = 35 x 10-3 L
Substituting these values in the above equation, we get
(a) Define the following terms :
(i) Molarity
(ii) Molal elevation constant (Kb)
(b) A solution containing 15 g urea (molar mass = 60 g mol–1) per litre of the solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol–1) in water. Calculate the mass of glucose present in one litre of its solution.
OR
(a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(b) A solution of glucose (molar mass = 180 g mol–1) in water is labelled as 10% (by mass). What would be the molality and molarity of the solution?
(Density of solution = 1.2 g mL–1)
(i) Molarity of a solution is defined as the number of moles of solute present in one litre of the solvent.
(ii) Molal elevation constant (Kb) is defined as the elevation of the boiling point of a solution when one mole of a non-volatile solute is dissolved in one kilogram of a volatile solvent.
(b) Osmotic pressure π = CRT = nVRT/2
where n is the number of moles of solute present in volume V of solvent.
It is given that the solution of urea is isotonic with the solution of glucose, thus
Or
a) A mixture of ethanol and acetone shows positive deviation from Raoult's Law. Pure ethanol possesses hydrogen bonding. The introduction of acetone between the molecules of ethanol results in breaking of some of these hydrogen bonds. Due to the weakening of interactions, the solution shows positive deviation from Raoult’s law.
(b)
10% by mass solution of glucose in water means that 10 g of glucose in present in 100 g of the solution i.e., 10 g of glucose is present in (100 - 10) g = 90 g of water.
Molar mass of glucose (C6H12O6) = 180 g mol-1 
Define the following terms:
Colligative properties
Colligative properties are properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present.Colligative properties include vapour pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.
Define the following terms:
Molality (m)
Molality of a solution is defined as the number of moles of solute dissolved per kg of solvent.
A 10% solution (by mass) of sucrose in water has the freezing point of 269.15 K.Calculate the freezing point of 10% glucose in water if freezing point of pure water is273.15 K.
Given : (Molar mass of sucrose = 342 g mol–1) (Molar mass of glucose = 180 g mol–1)
Given,
T1=269.15 K
T2 = 273.15
Molar mass of sucrose = 342 g mol–1
Molar mass of glucose = 180 g mol–1
ΔTf = Kf m
Here , m = w2x 1000/ M2XM1
273.15-269.15 = Kf x 10 x1000/ 342 x90
Kf = 12.3 K kg/mol
ΔTf = Kf m
= 12.3 x 10 x1000/ 180x90
= 7.6 K
Tf = 273.15 – 7.6 = 265.55 K
Write one difference in each of the following:
Solution and Colloid
| Solution | Colloid | |
| Appearance | It is clear, transparent and homogeneous | It is Cloudy but uniform and homogeneous |
| Tyndall effect | solution does not show Tyndall effect | colloid Shows Tyndall effect |
Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol–1) in 250 g of water.
(Kf of water = 1.86 K kg mol–1)
We know,
W1 = 250 g, w2 = 60 g, mw2 = 180 g/mol, Kf = 1.86 k kg mol–1
Give reasons for the following :
Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers.
Protein is the high molecular mass material the magnitude of colligative property depends inversely on the molecular mass and osmotic pressure.
Osmotic pressure is fine at room temp, no special arrangements are required and it depends on molarity whereas all other colligative properties depend on molality.
Give reasons for the following:
Aquatic animals are more comfortable in the cold water than in warm water.
The solubility of oxygen decreases as temperature rises which means solubility of oxygen in warm water is less as compared to cold water. Hence, aquatic animals are more comfortable in the cold water.
Give reasons for the following:
Elevation of the boiling point of 1 M KCl solution is nearly double than that of 1 M sugar solution.
The elevation of boiling point is Δtb = iKbm in both 1M KCl & 1 M sugar. As colligative property depends only on a number of particle and KCl as electrolyte will produce a double number of a particle of sugar. Hence the value van’t hoff factor is twice in 1m KCl then 1M sugar due to which elevation in boiling point is more.
18 g glucose (C6H12O6 ) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is:
76.0
752.4
759.0
7.6
B.
752.4
Vapour pressure of water (p°) = 760 torr
Number of moles of glucose
Molar mass of water = 18 g/mol
Mass of water (given) = 178.2 g
Number of moles of water
Total number of moles = (0.1 +9.9) = 10 moles
Now, the mole fraction of glucose in solution = Change in pressure with respect to initial pressure.
therefore, Vapour pressure of solution = (760-7.6)torr
=752.4 torr
The vapour pressure of acetone at 20oC is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20oC, its vapour pressure was 183 torr. The molar mass (g mol-1 ) of the substance is:
32
64
128
488
B.
64
Given,
po = 185 torr at 20oC
ps = 183 torr at 20oC
Mass of non-volatile substance,
m= 1.2 g
Mass of acetone taken = 100 g
As we have, 
putting the values, we get
3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is:
18 mg
36 mg
42 mg
54 mg
A.
18 mg
The initial strength of acetic acid = 0.06N
Final strength = 0.042 N
Volume given = 50 mL
there Initial m moles of CH3COOH
= 0.06 x 50 = 3
Final m moles of CH3COOH
= 0.042 x 50 = 21
therefore, m moles of CH3COOH absorbed
= 3-2.1
= 0.9 m mol
Hence, mass of CH3COOH absorbed per gram of charcoal
= 
Consider separate solution of 0.500 M C2H5OH (aq), 0.100 M Mg3(PO4)2 (aq) 0.250 M KBr (aq) and 0.125 M Na3PO4 (aq) at 25oC. Which statement is true about these solutions, assuming all salts to be strong electrolytes?
They all have the same osmotic pressure
0.100 M Mg3(PO4)2 (aq) has the highest osmotic pressure
0.125 M Na3PO (aq) has the highest osmotic pressure
0.500 M C2H5OH (aq) has the higest osmotic pressure
A.
They all have the same osmotic pressure
effective molarity = Van't Hoff factor x molarity
0.5 M C2H5OH (aq) i =1
Effective molarity = 0.5
0.25 M KBr (aq) i = 2
Effective molarity = 0.5
0.1 M Mg3(PO4)2 (aq) i = 5
Effective molarity = 0.5 M
0.125 M Na3PO4 (aq)
Effective molarity = 0.5 M
Hence, all colligative properties are same
The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2(M) HCl will be
0.875 M
1.00 M
1.75 M
0.0975 M
A.
0.875 M
M1V1 + M2V2 = MV
The density of a solution prepared by dissolving 120 g of urea (mol. Mass = 60 u ) in 1000g of water is 1.15 g/mL. The molarity of this solution is
0.50 M
1.78 M
1.02 M
2.05
D.
2.05
Kf for water is 1.86K kg mol–1.If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to –2.8°C?
72 g
93 g
39 g
27 g
B.
93 g
Coolant is glycol (C2H6O2) and is non-electrolyte.
ΔTf =2.8°
Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it form freezing at - 6° C will be : (Kf for water = 1.86 K kg mol-1, and molar mass of ethylene glycol = 62g mol-1)
804.32 g
204.30 g
400.00 g
304.60 g
A.
804.32 g
ΔTf = iKfm
ΔTf = 6ºC
w1 = mass of ethylene glycol in grams
w2 = mass of solvent (H2O) in grams= 4000g
m1 = Molar mass of ethylene glycol = 62 g
i = 1
i is van't off factor
A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is
1.8 atm
3 atm
0.3 atm
0.3 atm
A.
1.8 atm
Kp depends upon the partial pressure of reactants and products so first calculate their partial pressure and then, calculate Kp,
In the context of the lanthanoids, which of the following statement is not correct?
There is a gradual decrease in the radii of the members with increasing atomic number in the series.
All the member exhibit +3 oxidation state.
Because of similar properties, the separation of lanthanoids is not easy.
Availability of 4f electrons results in the formation of compounds in +4 state for all the members of the series.
D.
Availability of 4f electrons results in the formation of compounds in +4 state for all the members of the series.
If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water (ΔTf), when 0.01 mol of sodium sulphate is dissolved in 1 kg of water, is (Kf = 1.86 K kg mol–1).
0.0372 K
0.0558 K
0.0744 K
0.0186 K
B.
0.0558 K
Na2SO4 → 2Na+ + SO42-
Van't Hoff factor of dissociation for Na2SO3 = 3
Molality of the solution = 0.01 m
ΔTf = i x Kf x m
= 3 x 1.86 x 0.01
= 0.0558 K
On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively.Vapour pressure of the solution obtained by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane = 100 g mol–1 and of octane =114 g mol–1)
144.5 kPa
72.0 kPa
36.1 kPa
96.2 kPa
B.
72.0 kPa
Vapour pressure of the solution can be calculated by the formula
pT = XHp°H + Xo.P°o.
Therefore calculate the mole fractions of heptane and octane
pT = XHp°H + Xo.P°o.
The freezing point of benzene decreases by 0.45°C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be
(Kf for benzene = 5.12 K kg mol–1)
64.6%
80.4%
74.6%
94.6%
D.
94.6%
In benzene
2CH3COOH ⇌ (CH3COOH)2
On treatment of 100 mL of 0.1 M solution of CoCl3 . 6H2O with excess AgNO3; 1.2 × 1022 ions are precipitated. The complex is
[Co(H2O)4 Cl2]Cl.2H2O
[Co(H2O)3Cl3].3H2O
[Co(H2O)6]Cl3
[Co(H2O)5Cl]Cl2.H2O
D.
[Co(H2O)5Cl]Cl2.H2O
It means 2Cl– ions present in ionization sphere
∴ complex is [Co(H2O)5Cl]Cl2.H2O
1 gram of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2. The molar mass of M2CO3 in g mol-1 is
1186
84.3
118.6
11.86
B.
84.3
Given chemical equation
M2CO3 +2HCl → 2MCl + H2O + CO2
1 gm 0.01186 mol
From the balanced chemical equation
1/M = 0.01186
M = 84.3 gm/mol
A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statements is correct regarding the behaviour of the solution?
The solution formed is an ideal solution
The solution is non-ideal, showing +ve deviation from Raoult’s law.
The solution is non-ideal, showing –ve deviation from Raoult’s law.
n-heptane shows +ve deviation while ethanol shows –ve deviation from Raoult’s law.
B.
The solution is non-ideal, showing +ve deviation from Raoult’s law.
On mixing n-heptane and ethanol; strong interactions are replaced by weak interaction and hence it kes non-ideal solution with positive deviation.
At 80o C, the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture solution of ‘A’ and ‘B’ boils at 80o C and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg)
52 mol percent
34 mol percent
48 mol percent
50 mol percent
D.
50 mol percent
PT = PoXA + PoXB
760 = 520XA+ PoB(1-XA)
⇒ = 0.5
Thus, mole% of A = 50%
Equal masses of methane and oxygen are mixed in an empty container at 25ºC. The fraction of the total pressure exerted by oxygen is
2/3
(1/3) x (273 / 298)
1/3
1/2
C.
1/3
Let the mass of methane and oxygen is w
mole fraction of oxygen
Let the total pressure be P The pressure exerted by oxygen (partial pressure) = XO2 × Ptotal
P =1/3
A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol–1) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 gcm–3, molar mass of the substance will be –
90.0g mol–1
115.0g mol–1
105.0g mol–1
210.0 g mol–1
D.
210.0 g mol–1
Solutions with the same osmotic pressure are isotonic
Let the molar mass of the substance be M
π1 =C1RT=C2RT =π2
So, C1 = C2
As density of the solutions are same
The density (in g mL–1) of a 3.60 M sulphuric acid solution that is 29% H2SO4 (Molar mass = 98 g mol-) by mass will be
1.64
1.88
1.22
1.45
C.
1.22
Let the density of solution be ‘d’
Molarity of solution given = 3.6
i.e. 1 litre of solution contains 3.6 moles of H2SO4
or 1 litre of solution contains 3.6 × 98 gms of H2SO4
Since, the solution is 29% by mass.
100 gm solution contains 29 gm H2SO4
100/d ml l solution contains 29 gm of H2SO4.
1000 ml solution contains 3.6 × 98 gm of H2SO4
d = 1.22
A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be
350
300
360
700
A.
350
Let the vapour pressure of pure ethyl alcohol be P,
According to Raoult’s law 290 = 200 × 0.4 + P × 0.6
P = 290-80/0.6 = 350 mm Hg
6.02×1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is
0.001 M
0.1 M
0.02 M
0.01 M
D.
0.01 M
Avogadros number NA = 6.02 x 1023 = 1 mole
∴ 6.02 x 1020 molecules
= 0.001 mole in 100 mL (0.1 L solution)
∴ Molar concentration = mol/volume in L
=0.001/0.1
= 0.01 M
To neutralize completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H3PO3), the volume of 0.1 M aqueous KOH solution required is
10 mL
60 mL
40 mL
20 mL
C.
40 mL
H3PO3 is a dibasic acid (containing two ionizable protons attached to O directly)
H3PO3 ⇌ 2H+ + HPO42-
0.1 M H3PO3 = 0.2 N H3PO3
0.1 M KOH = 0.1 N KOH
N1V1 = N2V2
(KOH) = (H3PO3)
0.1 V1 = 0.2 x 20
V1 = 40 mL
Which of the following liquid pairs shows a positive deviation from Raoult’s law?
Water – hydrochloric acid
Acetone – chloroform
Water – nitric acid
Benzene – methanol
D.
Benzene – methanol
Water and hydrochloric acid; and water and nitric acid form miscible solutions. They show no deviation.
Which one of the following statements is false?
Raoult’s law states that the vapour pressure of a components over a solution is proportional to its mole fraction
Two sucrose solutions of same molality prepared in different solvents will have the same freezing point depression
The correct order of osmotic pressure for 0.01 M aqueous solution of each compound is BaCl2 > KCl > CH3COOH > sucrose
The osmotic pressure (π) = MRT, where M is the molarity of the solution
B.
Two sucrose solutions of same molality prepared in different solvents will have the same freezing point depression
(a)pA = XA pAo : true
(b) ΔTf =Kfm
Kf is dependent on solvent
Thus is dependent on solvent
Thus freezing points [= t(solvent)- ΔTf) are different.
(c) i = [1 + (y-1) x]
y = number of ions
x = degree of ionisation
i = 3 for BaCl2 x = 1 (strong electrolyte)
i = 2 for KCl x = 1 (strong electrolyte)
i = (1+x) for CH3COOH x<i = (for BaCl2)>KCl>CH3COOH>Sucrose
(d) π = imRT = = MRT ; true (if van't Hoff factor i = 1
For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?
[Co(H2O)3 Cl3].3H2O
[Co(H2O)6]Cl3
[Co(H2O)5Cl] Cl2.H2O
[Co(H2O)4Cl2]Cl.2H2O
A.
[Co(H2O)3 Cl3].3H2O
The complex giving least number of ions will have lowest depression in freezing point and therefore highest freezing point.(Van’t Hoff factor = 1)
An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1 M solution of Na2SO4 is added, BaSO4 just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4 is 1 x 10–10. What is the original concentration of Ba2+?
1.0 x 10–10 M
5 x 10–9 M
2x10–9 M
1.1 x 10–9M
D.
1.1 x 10–9M
For the calculation of [Ba+2] in original solution (450 ml)
Which of the following statements about the composition of the vapour over an ideal 1:1 molar mixture of benzene and toluene is correct? Assume that the temperature is constant at 250C.
(Given, vapour pressure data at 250C benzene = 12.8 kPa, toluene = 3.85 kPa)
The vapour will contain a higher percentage of toluene
The vapour will contain equal amounts of benzene and toluene
Not enough information is given to make a prediction
The vapour will contain a higher percentage of benzene
D.
The vapour will contain a higher percentage of benzene
A component having vapour pressure will have a higher percentage in the vapour phase. Benzene has vapour pressure 12.8 kPa which is greater than toluene 3.85pKa. Hence, the vapour will contain a higher percentage of benzene.
which is one not equal to zero for an ideal solution?
B.
ΔSmixFor an ideal solution
i) There will no change in volume on mixing the two components i.e. ΔVmixing =0
ii) There will no change in volume of ΔHmixing =0 so, ΔSmix not equal to for an ideal solution.
A mixture of gases contains H2 and O2 gases in the ratio of 1:4 (w/w).what is the molar ratio of the two gases in the mixture?
1:4
4:1
16:1
2:1
B.
4:1
Let the mass of H2 gas be x g and mass of O2 gas 4x g
Molar H2 : O2
mass 2 : 32
The boiling point of 0.2 mol kg-1 solution of X in water is greater than the equimolal solution of Y in water. Which one of the following statements is true is this case?
X is undergoing dissociation in water
The molecular mass of X is greater than the molecular mass of Y.
Molecular mass of X is less than the molecular mass of Y,
Y is undergoing dissociation in water while X undergoes no change.
A.
X is undergoing dissociation in water
Molality of solution X = molality of solution Y = 0.2
we know that elevation in the boiling point( ΔTb ) of a solution is proportional to the molal concentration of the solution i.e.
ΔTb = Kbm
where, m is the molality of the solution and Kb is molal boiling point constant or ebullioscopic constant.
therefore,
By elevation in boiling point relation
ΔTb = iKbm or ΔTb directly proportional to i
where i is van't Hoff is a factor
Since ΔTb of solution X is greater than ΔTb of solution Y.
(observed colligative property is greater than normal colligative property).
therefore, i of solution > i of solution Y
therefore solution X undergoing dissociation.
Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?
KCl
C6H12O6
Al2(SO4)3
K2SO4
C.
Al2(SO4)3
ΔTf (freezing point depression) is a colligative property and depends upon the van't Hoff factor (i) i,e., number of ions given by the electrolyte in aqueous solution.
i
The weight of silver (at. wt. = 108) displaced by a quantity of electricity which displaces 5600 mL of O2 at STP will be
5.4 g
10.8 g
54.0 g
108 g
D.
108 g
Since, 22400 mL volume is occupied by 1 mole of O2 STP.
Thus, 5600 mL of O2 means =
Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2) at 25oC are 200 mmHg and 41.5 mmHg respectively, Vapour pressure of the solution obtained by mixing 25.5 g of CHCl3 and 40 g of CH2Cl2 at the same temperature will be
(molecular mass of CHCl3= 119.5 u and molecular mass of CH2Cl2 = 85 u)
173.9 mmHg
615.0 mmHg
347.9 mmHg
28.5 mmHg
A.
173.9 mmHg
Number of moles of CHCl3
The van't Hoff factor, i for a compound which undergoes dissociation in one solvent and association in other solvent is respectively.
Less than one and less than one
Greater than one and less than one
Greater than one and greater than one
less than one and greater than one
B.
Greater than one and less than one
For dissociation, i > 1.
For association, i <1.
An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase?
Addition of NaCl
Addition of Na2SO4
Addition of 1.00 molal KI
addition of water
D.
addition of water
Vapour pressure depends upon the surface area of the solution. Larger the surface area, higher is the vapour pressure.
The addition of solute decreases the vapour pressure as some sites of the surface are occupied by solute particles, resulting in decreased surface area. However, the addition of solvent, ie, dilution increases the surface area of the liquid surface, thus results in increased vapour pressure.
Hence, the addition of water to the aqueous solution of (1 molal) KI, result in increased vapour pressure.
What is the [OH-] in the final solution prepared by mixing 20.0 mL of 0.50 M HCl with 30.0mL of 0.10 Ba(OH)2?
0.10 M
0.40 M
0.0050 M
0.12 M
A.
0.10 M
Number of milliequivalents of HCl = 2 x 0.050 x 1 = 1
Number of milliequivalents of Ba(OH)2 = 2 x 30 x 0.10 = 6
[OH]- of final soluton
milliequivalent of Ba(OH)2
A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)Cl freezes at -0.00732o C . Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (kf = - 1.86o C/m)
2
3
4
1
A.
2
Given,
molality = 0.0020 m
Δ Tf = 0o C -0.007320 C
kf = 1.86 oC/m
ΔTf = i.kf x m
i = ΔTf/ kf x m
= 0.00732/1.82 x 0.0020
= 1.92 = 2
A solution containing 10 g per dm3 is urea (molecular mass = 60 g mol-1) is isotonic with a molecular mass of this non-volatile solute. The molecular mass of this of this non-volatile solute is:
250 g mol-1
300 g mol-1
350 g mol-1
200 g mol-1
B.
300 g mol-1
10 g per dm3 of urea is isotonic with 5% solution of a non-volatile solute. Hence, between this solution osmosis is not possible so, their molar concentrations are equal to each other,
Thus, molar concentration of urea solution
1.00 g of a non- electrolyte solute (molar mass 250 g mol-1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K Kg mol-1, the freezing point of benzene will be lowered by:
0.4 K
0.3 K
0.5 K
0.2 K
A.
0.4 K
Molality of non- electrolyte solute
A solution of acetone in ethanol:
shows a negative deviation from Raoult's law
shows a positive deviation from Raoult's law
behave likea near ideal solution
obeys Raoult's law
B.
shows a positive deviation from Raoult's law
A solution ofacetone in ethanol shows a positive deviation from Raoult's law due to miscibility of these two liquids with a difference of polarity and length of the hydrocarbon chain.
During osmosis, flow of water through a semi-permeable membrane is:
from a solution having higher concentration only
from both sides of a semi-permeable membrane with equal flow rates
from both sides of a semi-permeable membrane with unequal flow rates
from a solution having lower concentration only
D.
from a solution having lower concentration only
During osmosis, the flow of water through a semi-permeable membrane is from a solution having lower concentration only.
If molality of the dilute solution is doubled, the value of molal depression constant (Kf) will be
Doubled
Halved
Tripled
Unchanged
D.
Unchanged
Kf (molal depression constant) is a characteristic of solvent and is independent of molality.
Which of the following is dependent on temperature?
Molality
Molarity
Mole fraction
Weight percentage
B.
Molarity
Molarity includes volume of solution which can change with change in temperature.
The normality of 10% (w/v) of acetic acid is
1 N
1.3 N
1.7 N
1.9 N
C.
1.7 N
Normality is given as,
Consider the following solutions,
A = 0.1 M glucose,
B = 0.05 M NaCl,
C = 0.05 M BaCl2
D = 0.1 M AlCl3
Which of the following pairs is isotonic?
A and B
A and D
A and C
B and C
A.
A and B
Isotonic solutions have a same molar concentration of solute particles in solution. Molar concentration of particles in solution are,
(a) = 0.1 M in glucose
(b) = 2 x 0.05 M in NaCl2
(c) = 3 x 0.05 M in BaCl2
(d) = 4 x 0.1 M in AlCl3
Hence, 0.1 M glucose and 0.05 and
Which of the following aqueous solution should have a highest boiling point?
1.0 M NaOH
1.0 M Na2SO4
1.0 M NH4NO3
10 MKNO3
B.
1.0 M Na2SO4
It depends on the Van't Hoff factor.
Van't Hoff factor is directly proportional to the boiling point.
In the case of Na2SO3, i is maximum = 3
Hence it has maximum boiling points.
Combustion of glucose takes place according to the equation.
C6H12O6 + 6O2 → 6CO2 + 6H2O
Δ = - 72 k-cal
The energy required for combustion of 1.6 g of glucose is
0.064 kcal
0.64 kCal
6.4 kcal
64 kcal
B.
0.64 kCal
Molar mass of C6H12O6 = 180 g mol-1
∵ 180 g of glucose require energy = 72 kcal
∴ 1.6 g of glucose require energy = 72 x 1.6/180
= 0.64 k cal
when the heat of reaction at constant pressure is -2.5 x 103 cal can entropy change is 7.4 cal deg-1 at 25oC, the reaction is predicted as
reversible
Spontaneous
non -spontaneous
Irreversible
B.
Spontaneous
For the reaction at constant pressure,
ΔH = - 2.5 x 103 cal
ΔS = 7.4 cal deg-1
∴ ΔG = ΔH -TΔS
= (-)2.5 x 103 - 298 x 7.4
= - 2.5 x 103 - 2205.2
ΔG = - 4705 cal
Hence, reaction is spontaneous.
The volume of water to be added to 100 cm3 of 0.5 NH2SO4 to get decinormal concentration is
100 cm3
450 cm3
500 cm3
400 cm3
D.
400 cm3
N1V1 = N2V2
Given, N1 = 0.5 N, N2 = 0.1 N
V1 = 100 cm3 (Total final volume)
Therefore, Volume to be added
= 500- 100 = 400 cm3
Sponsor Area
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