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_{A}X_{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*

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