Question

A mixture of chlorobenzene and water (which are virtually immiscible) boils at $90.3{}^{\circ }C$ at an external pressure of $740.2mm$. The vapour pressure of pure water at $40.3^{\circ }C$ is $530.1mm$. What is the percentage weight of chlorobenzene in the distillate?

• A. 20.21
• B. 37.19
• C. 28.74
• D. 71.26

Answer 1

Answer: (D)

Partial pressure of liquids are proportional to their molar concentrations.

$\frac{P_A}{P_B}=\frac{N_A}{N_B}=\frac{W_A/M_A}{W_B/M_B}$

$P_A$ and $P_B$ are the partial pressures of two immiscible liquids $A$ and $B$.

Similarly, partial pressure of chlorobenzene,

$P_c=740.2-530.1=210.1\cdot mm$

Partial pressure of water $=P_w=530.1mm$

$\frac{P_c}{P_w}=\frac{N_c}{N_w}=\frac{W_c\times M_w}{W_w\times M_c}=\frac{W_c\times 18}{W_w\times 112.5}$

$\frac{P_c}{P_w}=\frac{210.1}{530.1}=\frac{W_c\times 18}{W_w\times 112.5}$

$\Rightarrow \frac{W_c}{W_w}=\frac{210.1\times 112.5}{530.1\times 18}=2.48$

$\frac{W_c}{W_w}=2.48$ or $W_c=2.48\times W_w$

or $W_c+W_w=3.48\times W_w$

$\%W_c=\frac{2.48\times W_w}{3.48\times W_w}\times 100=71.26$