Question

A certain aqueous solution of a low-molecular-weight solid nonelectrolyte freezes at -0.64 ^oC. For this solution, estimate (a) the normal boiling point; (b) the vapor pressure at 25 ^oC. The vapor pressure of pure water is 23.76 torr at 25 ^oC.

K_f= 1.860 Kkg/mol. K_b= 0.513 ^oCkg/mol

Answer 1

Given depression in freezing point = 0.64 and Kf = 1.86 K Kg/mol

   the relation between tese two is delta Tf = Kf .molality

hence molality of solution = Tf/Kf = 0.64/1.86 = 0.344

Q1) calculation of normal B.P. of solution

    delta Tb = Kb . molality

               = 0.513 x 0.344 = 0.1765 K

Thus the normal boiling point of solution is 100 + 0.1765 = 100.1765 ^o C

Q2) calculation of vapor pressurre

   We have the relation according to Raoults law as mole fraction of solute (X_B) = (P^o-P)/P^o where P^o is the vapor pressure of pure solvent and P is the vapor pressure of solution.

   molality of a solution is number of moles of solute in one Kg of solvent by definition.

one Kg of water has number of moles = 1000g/18gmol^-1 = 55.5 and number of moles of solute = molality =0.344

Mole fraction = moles of solute / total moles = 0.344/(55.5 + 0.344)= 0.00616

      From Raoult's law 0.00616 = (23.76 - x)/ 23.76 and solving for x we get

         x = 23.61

thus the vapor pressure of solution at 25^o C is 23.61 torr.