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. 79
• D. 38.46

Answer 1

The correct option is C 79

Vapour pressure = $7.7\times {10}^4$ Pa

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

Partial pressure = Vapour pressure = Mole fraction x 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

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

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

Weight percent of chloroform = Mole fraction of $CHC{l}_3$ x Molar mass of $CHC{l}_3$ / [Molar mass of $CHC{l}_3$ +Molar mass of $H_{2}O$ }

Weight percent of chloroform = 79%

Answer is 79%.