Water and chlorobenzene are immiscible liquids. Their mixture boils at 89^oC under a reduced pressure of 7.7 x 10⁴ Pa. The vapour pressure of pure water at 89^oC is 
7 x 10⁴ pa. Weight per cent of chlorobenzene in the distillate is :

Given:

• Boiling point of the mixture: 89°C
• Total pressure (boiling point pressure of the mixture): P_total = 7.7 × 10⁴ Pa
• Vapor pressure of pure water at 89°C:P_water = 7 × 10⁴ Pa
• Vapor pressure of chlorobenzene: To be calculated

We know that the total vapor pressure of the mixture is the sum of the individual vapor pressures:

P_total = P_water + P_{chlorobenzene}

Substituting the given values:

7.7 × 10⁴ = 7 × 10⁴ + P_{chlorobenzene}

Therefore:

P_{chlorobenzene} = 7.7 × 10⁴ - 7 × 10⁴ = 0.7 × 10⁴ Pa

From Raoult's Law, we know that:

χ_{chlorobenzene} = P_{chlorobenzene} / P_total

Substituting the values:

χ_{chlorobenzene} = (0.7 × 10⁴) / (7.7 × 10⁴) = 0.0909

Using the formula for weight percent:

Weight percent of chlorobenzene = (χ_{chlorobenzene} × M_{chlorobenzene}) / [(χ_{chlorobenzene} × M_{chlorobenzene}) + (χ_water × M_water)] × 100

Where:

• M_{chlorobenzene}: Molar mass of chlorobenzene = 112 g/mol
• M_water: Molar mass of water = 18 g/mol
• χ_water: Mole fraction of water = 1 - 0.0909 = 0.9091

Substituting the values:

Weight percent of chlorobenzene = (0.0909 × 112) / [(0.0909 × 112) + (0.9091 × 18)] × 100Weight percent of chlorobenzene = 38.46%

Final Answer: Option (d) 38.46%