Question

The vapor pressure of pure liquid water is p*(H2O) = 0.02308 atm. Above an aqueous solution with 122 g of a non-volatile solute (molar mass 241 g/mol) in 920 g H2O has a vapor pressure of 0.02239 atm. Calculate the activity and the activity coefficient of water in the solution.

Answer 1

Ans:

The vapor pressure of pure liquid water is p*(H2O) = 0.02308 atm. Above an aqueous solution with 122 g of a non-volatile solute (molar mass 241 g/mol) in 920 g H2O has a vapor pressure of 0.02239 atm.

Moles solute = n_B= 122 g (1. mol/241. g) = 0.5062 mol

Moles water = n_A = 920 g (1. mol/18.01 g) = 51.083 mol

X_B= n_B/(n_A+ n_B) = (0.5062)/(51.083 + 0.5062) = 0.00981 and X_A= 0.99019

Activity of water in the solution a_A = p_H2O/p_H2O* = 0.02239/0.02308 = 0.9701

Since a_A= g_AX_A, then g_A= a_A/X_A

Therefore Activity coefficient of water of water in the solution = g_A = 0.9701/0.99019 = 0.980