Question

How many grams of non-volatile compound b (molar mass= 97.80 g/mol) would need to be added to 250.0 g of water to produce a solution with a vapor pressure of 23.756 torr? The vapor pressure of water at this temperature is 42.362 torr.

Answer 1

Answer:1,063.12 grams of volatile compound should be added.

Explanation:

Vapor pressure of pure water =$p^o$ = 42.362 torr

Vapor pressure of solution =$p_s$ = 23.756 torr

Moles of water = $n_1=\frac{250 g}{18 g}=13.88 mol$

Moles of compound b=$n_2=\frac{x}{97.80 g/mol}$

$\frac{p^o-p_s}{p^o}=\frac{n_2}{n_1+n_2}$

$\frac{42.362 torr-23.756 torr}{42.362 torr}=\frac{\frac{x}{97.80 g/mol}}{\frac{x}{97.80 g/mol}+13.88 mol}$

x = 1,063.12 g g

1,063.12 grams of volatile compound should be added.