CBSE Class 12 Chemistry 2016 Delhi Set 1 Solved Paper

Question : 26

Total: 26

(a) Calculate the freezing point of solution when 1.9g of MgCl2(M=95gmol−1) was dissolved in 50g of water, assuming MgCl2 undergoes complete ionization. (Kf. for water =1.86K‌kg mol−1 )
(b) (i) Out of 1M glucose and 2M 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 2.56g of sulphur was dissolved in 100g of CS2, the freezing point lowered by 0.383K. Calculate the formula of sulphur (SX). (Kf. for CS2=3.83K‌kg‌mol−1, Atomic mass of Sulphur =32gmol−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 ?.

Solution:

(a) ‌WB(‌ solution ‌)=1.9g‌ws
‌W(H2O)=50g
‌M(MgCl2)=95g∕mol
‌i=3(‌ for ‌MgCl2.‌ ) ‌
‌m=‌

WB‌ in ‌kg

‌ Molar mass ‌×‌ mass of solvent ‌

‌∆Tf=iKfm
‌∆Tf=3×1.86×‌

1.9

×‌

1000

(kg)
‌∆Tf=2.232K
‌∆Tf=∆Tf∘−Tf
‌‌ Also, ‌‌‌∆Tf=∆Tf∘−Tf
‌Tf=Tf∘−∆Tf=273.15−2.232
‌=270.918K
(b) (i) 2M glucose, because more will be the concentration molality, more will be the elevation in boiling point.
(ii) Reverse osmosis takes place.
OR
(a) ∆Tf‌=iKfm
∆Tf‌=Kf×‌

WA(Kg)

×‌

(Kg)
MB‌=‌

∆Tf×WA(Kg)

‌=‌

3.83×2.56×1000

0.383×100(kg)

‌=256g∕mol
MB‌=n×‌ Atomic mass ‌
n‌=‌

‌ M ‌

‌=‌

256

=8
∴S8
(b) (i) Water moves out from blood cell, hence will shrink.
(ii) Water will enter into blood cell, hence will swell.

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