Question

When the mixture of two immiscible liquids (water and nitrobenzene) boils at $372K$ and the vapour pressure at this temperature are $97.7kPa\left(H_{2}O\right)$ and $3.6kPa\left(C_6{H}_{5}N{O}_2\right)$. Calculate the weight $\%$
of nitrobenzene in the vapour.

Answer 1

For two immiscible liquids
$\frac{w_1}{w_2}=\frac{P_1^{\circ }{M}_1}{P_2^{\circ }{M}_2}=\frac{3.6\times 123}{97.7\times 18}=0.2518$
$\frac{W_2}{W_1}=3.971$
$\frac{W_2+w_1}{W_1}=4.971$
$\frac{W_1}{W_2+W_1}\times 100=20.11\%$