Solutions

Sponsor Area

Question
CBSEENCH12011360

0.5 Molal aqueous solution of aweak acid (HX) is 20% ionised. If Kf for water is1.86 K kg mol-1, the lowering in freezing point of the solution is:

  • -1.12 K 

  • 0.56 K 

  • 1.12 K 

  • -0.56 K 

Solution

C.

1.12 K 

stack stack HX space with weak space acid below with 1 minus alpha below space space rightwards harpoon over leftwards harpoon space space space stack stack space straight H to the power of plus with 0 below space with straight alpha below plus stack space stack straight X to the power of minus end exponent with 0 below with straight alpha space below

straight alpha space equals space 20 percent sign space dissociation
space straight i space equals space 1 space minus space straight alpha space plus space straight alpha space plus straight alpha
space equals space 1 space plus space straight alpha space equals space 1 space plus space 0.2 space equals space 1.2
increment straight T subscript straight f space equals space straight i space straight x space straight K subscript straight f space straight x space straight m
space equals space 1.2 space straight x space 1.86 space straight K space kg space mol to the power of negative 1 end exponent space straight x space 0.5
space equals space 1.12 space straight K

Sponsor Area

Question
CBSEENCH12011403

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

Solution

A.

0.4 K

Molality of non- electrolyte solute

fraction numerator begin display style fraction numerator weight space of space solute space in space gram over denominator molecular space weight space of space solute end fraction end style over denominator weight space of space solvent space in space kg end fraction

equals space fraction numerator begin display style 1 over 250 end style over denominator 0.0512 end fraction space equals space fraction numerator 1 over denominator 250 space straight x space 0.0512 end fraction space equals space 0.0781 space straight m

increment straight T subscript straight f space equals space straight k subscript straight f space straight x space molality space of space solution
equals space 5.12 space straight x space 0.0781 space equals space 0.4 space straight K

Question
CBSEENCH12011244

200 mL of an aqueous solution of a protein contains its 1.26 g . The osmotic pressure of this solution at 300 K is found to be  2.57 x 10-3 bar. The molar mass of protein will be (R = 0.083 L bar mol-1 K-1)

  • 51022 g mol-1

  • 122044 g mol-1

  • 31011 g mol-1

  • 61038 g mol-1

Solution

D.

61038 g mol-1

straight pi space equals space CRT

equals space fraction numerator straight w space straight x space 1000 over denominator straight M space straight x space straight V space left parenthesis in space mL right parenthesis end fraction space straight x space RT

open square brackets therefore space space straight C space equals space fraction numerator straight n space straight x space 1000 over denominator straight V space in space mL end fraction and space straight n space equals space straight w over straight M close square brackets

equals space fraction numerator 1.26 space straight x space 1000 space straight x space 0.083 space straight x space 300 over denominator 2.57 space straight x space 10 to the power of negative 3 end exponent space straight x space 200 end fraction

equals space 61038 space straight g space mol to the power of negative 1 end exponent

Question
CBSEENCH12011284

25.3 g of sodium carbonate, Na2CO3 is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely molar concentration of sodium ion, Na+ and carbonate ion, CO32- are respectively (Molar mass of Na2CO3 = 106 g mol-1)

  • 0.955 M and 1.910 M

  • 1.910 M and 0.955 M 

  • 1.90 M and 1.910 M 

  • 0.477 M and 0.477 M 

Solution

B.

1.910 M and 0.955 M 

Molarity space equals space fraction numerator number space of space moles space space of space solute over denominator volume space of space solution space left parenthesis in space mL right parenthesis end fraction straight x space 1000
space
equals space fraction numerator 25.3 space straight x space 1000 over denominator 106 space straight x space 250 end fraction space equals space 0.9547 space equals space 0.955 space straight M

Na subscript 2 CO subscript 3 space in space aqueous space solution space remains space dissociated space as

stack Na subscript 2 CO subscript 3 space space with straight x below rightwards arrow space stack 2 Na to the power of plus space with 2 straight x below plus space stack CO subscript 3 superscript 2 minus end superscript with straight x below
Since comma space the space molarity space of space Na subscript 2 CO subscript 3 space is space 0.955 space straight M comma space the space molarity space of space CO subscript 3 superscript 2 minus end superscript space
is space also space 0.955 space straight M space and space that space of space Na to the power of plus space is space 2 space straight x space 0.955 space equals space 1.910 space straight M

Question
CBSEENCH12011324

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

Solution

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

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