' W ' g of a non-volatile electrolyte solid solute of molar mass ' M ' g mol ^ -1 when dissolved in 100 mL water, decreases vapour pressure of water from 640 mm Hg to 600 mm Hg. If aqueous solution of the electrolyte boils at 375 K and K _ b for water is 0.52 ~K ~kg ~mol ^ -1 , then the mole fraction of the electrolyte solute (x_ 2 ) in the solution can be expressed as (Given : density of water =1 ~g / mL and boiling point of water =373 ~K )
D. 1 3 8 W M
Given the vapour pressure of pure water P^o = 640 mm Hg and vapour pressure of solution P_s = 600 mm Hg. According to Raoult's law for an electrolyte solution, the relative lowering of vapour pressure is given by P^o - P_s P^o = i x_2, where i is the van't Hoff factor and x_2 is the mole fraction of the solute. Substituting the values: 640 - 600 640 = i x_2 40 640 = i x_2 i x_2 = 1 16 ... (1) The elevation in boiling point is given by T_b = i K_b m, where m is the molality. Given T_b = 375 K and T_b^o = 373 K, so T_b = 375 - 373 = 2 K. Molality m = W M 100 1000 = 10W M (since mass of water = 100 mL 1 g/mL = 100 g). Substituting in the elevation formula: 2 = i 0.52 10W M i = 2M 5.2W = 20M 52W = 5M 13W ... (2) Substitute the value of i from (2) into (1): 5M 13W x_2 = 1 16 x_2 = 1 16 13W 5M = 13 80 W M This can be rewritten as x_2 = 1.3 8 W M .
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