Question

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

• A. 10 %
• B. 28.5 %
• C. 72 %
• D. 20.1 %

Answer 1

The correct option is D 20.1 %

Since the two liquids given are immiscible, the mole fraction for both in liquid mixture will be taken as 1
total vapour pressure of liquid mixture $=97.7+3.6=101.3kPa$
Now mole fraction of nitrobenzene in vapour phase $=\frac{3.6}{101.3}=0.0355$
Now weight percentage of $N{O}_2$ in vapour phase $=\frac{0.0355\times 123}{(1-0.0355)\times 18+0.0355\times 123}\times 100$
=20.1%