Question

A mixture of chlorobenzene and wate (immiscible) boils at ${90.3}^{o}C$ at an external pressure of 740.2 mm. The vapour pressure of pure water at ${90.3}^{o}C$ is 530.1 mm. Calculate the % composition of distillate :

• A. $H_{2}O=$ 35%
• B. $H_{2}O=$ 22%
• C. $H_{2}O=$ 29%
• D. $H_{2}O=$ 71%

Answer 1

The correct option is C $H_{2}O=$ 29%

At boiling point $P_M^{\prime }=740.2mm$
$P_{H_{2}O}^{\prime }=530.1mm$
$\therefore P_{chlorobenzene}^{\prime }=740.2-530.1$
$=210.1mm$
Also, $P^{\prime }=P_M\times$ mole fraction
$530.1=740.2\times$ mole fraction of $H_{2}O$
$\therefore$ mole fraction of $H_{2}O=0.716$
$\therefore$ mole fraction of $C_6{H}_{5}Cl=0.284$
Let a g $H_{2}O$ and b g chlorobenzene be present in distillate
or $\frac{w_{H_{2}O}/18}{\frac{w_{CB}}{112.5}+\frac{W_{H_{2}O}}{18}}=0.716$ ....(i)
and $\frac{w_{H_{2}O}/112.5}{\frac{w_{CB}}{112.5}+\frac{W_{H_{2}O}}{18}}=0.284$ ....(ii)
By eq. (i)/(ii)
or $\frac{a}{18}\times \frac{112.5}{b}=\frac{0.716}{0.284}$
or $\frac{a}{b}=\frac{12.89}{31.5}=0.403$ .....(iii)
Let total distillate be 100 g
$\therefore a+b=100$ .....(iv)
By eqs. (iii) and (iv)
$b=$ 71.27%
$a=$ 28.73%