Question

Vapour pressure of $C_6{H}_6$ and $C_7{H}_8$ are 119 mm and 37 mm of Hg. Calculate molar composition of $C_6{H}_6$ and $C_7{H}_8$ in a mixture having V.P. of 80 mm of Hg. Also calculate vapour phase composition over the mixture:

• A. Liquid phase 52.44%, 47.56% vapour phase 78%, 22%
• B. Liquid phase 78%, 22% vapour phase 52.44%, 47.56%
• C. Liquid phase 60.45%, 39.50% vapour phase 82%, 18%
• D. Liquid phase 82%, 18% vapour phase 60.45%, 39.50%

Answer 1

The correct option is A Liquid phase 52.44%, 47.56% vapour phase 78%, 22%

Raoult's Law states that the partial vapour pressure of a component in a mixture is equal to the vapour pressure of the pure component at that temperature multiplied by its mole fraction in the mixture.

In a mixture of liquids of A and B,

$p_A=x_A\times {p_A}^0$

$p_B=x_B\times {p_B}^0$

Here, $p_A$ and $p_B$ are the partial vapour pressures of the components A and B. In any mixture of gases, each gas exerts its own pressure. This is called its partial pressure and is independent of the other gases present. Even if you take all the other gases away, the remaining gas would still be exerting its own partial pressure.

The total vapour pressure of the mixture is equal to the sum of the individual partial pressures.

$p=p_A+p_{B}p=x_A.{p_A}^0+x_B.{p_B}^0$

Given,

${p_A}^0=119$ mm of Hg

${p_B}^0=37$ mm of Hg

Resultant pressure is $p=80$ mm of Hg

Let mole fraction of A be $x_A$ and mole fraction of B will be $1-x_A$.

Then,

$80=119\times x_A+37(1-x_A)⟹x_A=0.524$

Hence, the solution contains $52.4\%$ of liquid A and $47.5\%$ of liquid B.

In vapour phase,

Pressure due to liquid A is $119\times 0.524=61.88$

Pressure due to liquid B is $37\times 0.47=17.39$

Composition of A in vapour phase will be $\frac{61.88}{(61.88+17.39)=}0.779$

Composition of B in vapour phase will be $\frac{17.39}{(61.88+17.39)}=0.221$

Hence, the vapour contains $77.9\%$ of vapour A and $22.1\%$ of vapour B.