CBSE Class 12 Chemistry 2017 Outside Delhi Set 1 Solved Paper

Question : 25

Total: 26

(a) A 10% solution (by mass) of sucrose in water has a freezing point of 269.15K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273.15K.
Given :
(Molar mass of sucrose =342gmol−1 )
( Molar mass of glucose =180gmol−1 )
(b) Define the following terms :
(i) Molality ( m )
(ii) Abnormal molar mass
OR
(a) 30g of urea (M=60gmol−1) is dissolved in 846g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298K is 23.8‌mm‌Hg.
(b) Write two differences between ideal solutions and non-ideal solutions.

Solution:

(a) ‌∆Tf=Kfm
‌‌ Here, ‌m=w2×1000∕M2×M1
‌273.15−269.15‌‌=Kf×10 ×1000∕342×90
‌Kf=12.3K‌kg∕mol
‌∆Tf=Kfm
‌‌=12.3×10×1000∕180×90
‌‌=7.6K
‌Tf=273.15−7.6=265.55K
(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)

(PA0−PA)

PA0

‌ =

(wB×MA)

(MB×wA)

‌

23.8−PA

23.8

‌=

(30×18)

×846
‌23.8−PA=23.8×[

(30×18)

×846]
‌23.8−PA=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) ∆mixH=0	(b) ∆mixH is not equal to 0 .
(c) ∆mixV=0	(c) ∆mixV is not equal to 0 .

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