Question

# The vapor pressure of pure liquid water is p*(H2O) = 0.02308 atm. Above an aqueous solution...

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.

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 = nB= 122 g (1. mol/241. g) = 0.5062 mol

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

XB= nB/(nA+ nB) = (0.5062)/(51.083 + 0.5062) = 0.00981 and XA= 0.99019

Activity of water in the solution aA = pH2O/pH2O* = 0.02239/0.02308 = 0.9701

Since aA= gAXA, then gA= aA/XA

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

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