Question

A mixture of chlorobenzene and water (which are virtually immiscible) boils at $90.3{ }^{\circ} C$ at an external pressure of $740.2\, mm$. The vapour pressure of pure water at $40.3^{\circ} C$ is $530.1\, mm$. 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 $=P_{w}=530.1 mm$

$\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}$

$\% W_{c}=\frac{2.48 \times W_{w}}{3.48 \times W_{w}} \times 100=71.26$