Solutions

Solutions

Question

(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)

Answer

(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
straight n subscript left parenthesis glucose right parenthesis end subscript VRT space space equals space straight n subscript urea VRT

rightwards double arrow nVRT space equals space 1560 space straight x space VRT
rightwards double arrow straight w subscript glucose space straight x space straight M subscript glucose space equals space 1560
straight w subscript glucose space equals space 1560 over straight M subscript glucose space

straight w subscript glucose space equals space 1560 over 180 space equals 45 space straight g
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-1Then comma space number space of space moles space of space glucose space equals space 10 over 180 mol space equals space 0.056 space mol
therefore space Molality space of space solution space equals space fraction numerator 0.056 space mol over denominator 0.09 space kg end fraction space equals space 0.62 space straight m
if space the space density space of space the space solution space is space 1.2 space straight g space mL to the power of negative 1 end exponent comma space then space the space volume space of space the space 100 space straight g space solution space can space be space given space as colon
fraction numerator 100 space straight g over denominator 1.2 space mL to the power of negative 1 end exponent end fraction equals 83.33 mL
equals 83.33 space straight x space 10 to the power of negative 3 end exponent space straight L

therefore space Molarity space of space the space solution space equals space fraction numerator 0.056 space mol over denominator 83.33 space straight x space 10 to the power of negative 3 end exponent space straight L end fraction space equals space 0.67 space straight M

 

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