Chemistry - Liquid Solution Question with Solution | TestHub

For immiscible liquids:

$P_{total}=P_{H_{2}O}+P_{NB}=97.7+3.6=101.3kPa$

Mole fraction of nitrobenzene in vapor:

$y_{NB}=\frac{3.6}{101.3}=0.03554$

Molar masses:

$M_{NB}=123$, $M_{H_{2}O}=18$

Weight % of nitrobenzene:

$$\frac{0.03554\times 123}{(0.03554\times 123)+(0.96446\times 18)}\times 100$$

$$=\frac{4.37}{4.37+17.36}\times 100=20.11\%$$

Therefore, the weight % of nitrobenzene in vapor = 20.11%