Q. 26 A
A. Calculate the freezing point of solution when 1.9 g of 
was dissolved in 50 g of water, assuming 
undergoes complete ionization.
B. (i) Out of 1 M glucose and 2 M 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?
Answer :
Kf = 1.86 K kg/mol
Mass of solute = 1.9g
Mass of solvent = 50g
Hence, Molality of the solution, 
MgCl2 undergoes complete ionization and yields 3 moles of constituent ions for every mole of MgCl2. Therefore i=3
Depression in freezing point, ΔTf = iKfm
= 3 x 1.86 x 0.4
= 2.232 K
Tf = 273.15 – 2.232 = 270.918 K
B. (i) The boiling point of a solution depends on the number of particles of the solute. The amount of solute is higher in 2M glucose solution as compared to 1M glucose solution and therefore 2M glucose solution has a higher boiling point than 1M glucose solution.
(ii) When the external pressure applied becomes more than the osmotic pressure of the solution, the process of reverse osmosis occurs in which the pure solvent starts flowing out of the solution through the semi permeable membrane.
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Match the terms given in Column I with expressions given in Column II.
