Given information:
- Water and Chlorobenzene are immiscible liquids.
- Their mixture boils at 89C under a reduced pressure of 7.7104 Pa.
- The vapour pressure of pure water at 89C is 7104 Pa.

To find: Weight per cent of Chlorobenzene in the distillate.

Solution:
1. Calculation of the total pressure in the system:
The reduced pressure given is the pressure at which the mixture boils. To find the total pressure in the system, we need to add the vapour pressure of pure water at 89C to the reduced pressure.
Total pressure = Vapour pressure of pure water + Reduced pressure
Total pressure = 7104 Pa + 7.7104 Pa
Total pressure = 7111.7104 Pa

2. Calculation of the composition of the vapour:
We assume that the vapour in equilibrium with the liquid mixture is composed of water and chlorobenzene in the same proportion as they exist in the liquid mixture. Let x be the weight per cent of chlorobenzene in the liquid mixture, then (100-x) is the weight per cent of water in the liquid mixture.

The vapour above the liquid mixture will also have the same composition as the liquid mixture.

Now, using Dalton's law of partial pressures, we can write:
Partial pressure of water in the vapour = mole fraction of water in the vapour x Total pressure
Partial pressure of chlorobenzene in the vapour = mole fraction of chlorobenzene in the vapour x Total pressure

We assume that the mole fraction of water in the vapour is equal to the mole fraction of water in the liquid mixture, and similarly for chlorobenzene.

Let's denote the mole fraction of water in the liquid mixture as y. Then, the mole fraction of chlorobenzene in the liquid mixture is (1-y).

Using the above assumptions and Dalton's law, we get:
Partial pressure of water in the vapour = y x 7111.7104 Pa
Partial pressure of chlorobenzene in the vapour = (1-y) x 7111.7104 Pa

3. Calculation of the weight per cent of chlorobenzene in the distillate:
We know that the mixture boils at 89C under the given reduced pressure. This means that the vapour pressure of the liquid mixture at 89C is equal to the reduced pressure.

Using Raoult's law, we can write:
Vapour pressure of the liquid mixture = mole fraction of water in the liquid mixture x Vapour pressure of pure water at 89C
+ mole fraction of chlorobenzene in the liquid mixture x Vapour pressure of chlorobenzene at 89C

Since water and chlorobenzene are immiscible, the mole fraction of each component in the liquid mixture is equal to its weight per cent in the liquid mixture.

At the boiling point, the vapour pressure of the liquid mixture is equal to the total pressure in the system. Therefore, we can write:
Total pressure = y x 7104 Pa + (1-y) x Vapour pressure of chlorobenzene at 89C

Solving for y, we get:
y = (7104 Pa - 7.7104 Pa)/(7104 Pa - Vapour pressure of chlor