CBSE Class 12 Chemistry 2017 Outside Delhi Set 1 Solved Paper

Question : 25 of 26

Marks: +1, -0

(a) A

10 \%

solution (by mass) of sucrose in water has a freezing point of

269.15 {\K}

. Calculate the freezing point of

10 \%

glucose in water if the freezing point of pure water is

273.15 {\K}

.
Given :
(Molar mass of sucrose

=342 {\g} {\mol}^{-1}

)

(

Molar mass of glucose

=180 {\g} {\mol}^{-1}

)
(b) Define the following terms :
(i) Molality (

m

)
(ii) Abnormal molar mass
OR
(a)

30 {\g}

of urea

( {\M}=60 {\g} {\mol}^{-1})

is dissolved in

846 {\g}

of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at

298 {\K}

23.8 {\mm} {\Hg}

.
(b) Write two differences between ideal solutions and non-ideal solutions.

Solution:

(a)

\;∆ {\T}_f= {\K}_f {\m}

\;\;\text " Here, "\; m=w_2 × 1000 \/ {\M}_2 × {\M}_1

\;273.15-269.15 \;\; = {\K}_f × 10

× 1000 \/ 342 × 90

\; {\K}_f=12.3 {\K} {\kg} \/ {\mol}

\;∆ {\T}_f= {\K}_f {\m}

\;\; =12.3 × 10 × 1000 \/ 180 × 90

\;\; =7.6 {\K}

\; {\T}_f=273.15-7.6=265.55 {\K}

(or any other correct method)
(b) (i) Number of moles of solute dissolved in per kilogram of the solvent.

(ii) Abnormal molar mass : If the molar mass calculated by using any of the colligative properties to be different than theoretically expected molar mass.
OR
(a) (i)

( {\P}_{ {\A}}^0- {\P}_{ {\A}}) / {\P}_{ {\A}}^0\;

= (w_{ {\B}} × {\M}_{ {\A}}) / ( {\M}_{ {\B}} × w_{ {\A}})

\;{23.8- {\P}_{ {\A}}}/{23.8}\;=(30 × 18) / 60 × 846

\;23.8- {\P}_{ {\A}}=23.8 ×[(30 × 18) / 60 × 846]

\;23.8- {\P}_{ {\A}}=0.2532

(b)

Ideal solution	Non-ideal solution
(a) It obeys Raoult's law over the entire range of concentration.	(a) Does not obeys Raoult's law over the entire range of concentration.
(b) $∆_{m i x} {\H}=0$	(b) $∆_{m i x} {\H}$ is not equal to 0 .
(c) $∆_{m i x} {\V}=0$	(c) $∆_{m i x} \V$ is not equal to 0 .

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