Question

Water and chlorobenzene are immiscible liquids. Their mixture boils at ${89}^{o}C$ under a reduced pressure of $7.7\times {10}^{4}Pa$. The vapour pressure of pure water at ${89}^{o}C$ is $7\times {10}^{4}Pa$. Weight percent of chlorobenzene in the distillate is:

• A. $50$
• B. $60$
• C. $78.3$
• D. $38.46$

Answer 1

Answer: (D)

$38.46$

Vapour pressure of $H_{2}O=7.7\times {10}^4$$

Total pressure$=7\times {10}^4$

Now, Partial pressure=Vapour pressure=Mole fraction$\times$ Total pressure

Mole fraction of $H_{2}O=\frac{Vapourpressure}{Totalpressure}=\frac{7.7\times {10}^4}{7\times {10}^4}=0.909$

Mole fraction of $CHC{L}_3=1-0.909=0.091={\chi }_{CHC{L}_3}$

Now, Molar mass of $H_{2}O=M_{H_{2}O}=18g/mol$

Molar mass of $CHC{L}_3=M_{CHC{L}_3}=119.5g/mol$

Now, Weight percent of chlorobenzene$={\chi }_{CHC{L}_3}\times \frac{M_{CHC{L}_3}}{M_{CHC{L}_3}+M_{H_{2}O}}=0.909\times \frac{119.5}{119.5+18}=78.26$