Question

The mole fraction of $CCl _{4}( g )$ in the vapour in equilibrium with liquid mixture of $CCl _{4}$ and $SiCl _{4}$ is $0.3474 .$ The vapour pressure of $SiCl _{4}$ and $CCl _{4}$ is $238.3$ and $114.9 \,mm$, respectively, at the same temperature. Calculate $\%$ by weight of $CCl _{4}$ in liquid mixture.

• A. 50 %
• B. 40 %
• C. 30 %
• D. 60 %

Answer 1

Answer: (A)

$X _{ CCl _{4}}( g )=0.3474$

$X _{ SiCl _{4}}( g )=0.6526$

$P _{ SiCl _{4}( g )}^{\circ}=238.3\, mm$

$P _{ CCl _{4}}^{\circ}=114.9 \,mm$

$P _{ CCl _{4}}'= P _{ CCl _{4}}^{\circ} \cdot X _{ CCl _{4}(\ell)}= P _{ M } \times X _{ CCl _{4}( g )} $

$\therefore 144.9 \times X _{ CCl _{4}(\ell)}= P _{ M } \times 0.3474 \dots$ (i)

Also, $ 238.3 \times X _{ SiCl _{4}(\ell)}= P _{ M } \times 0.6526 \dots$(ii)

By equations (i) and (ii), we get

$\frac{X_{ CCl _{4}(\ell)}}{X_{ SiCl _{4}(\ell)}}$

$=\frac{0.3474}{0.6526} \times \frac{238.3}{114.9}=1.104$

Let $a \,g \,CCl _{4}$ and $b\,g\, SiCl _{4}$ be present in the liquid and the ratio of mole fraction is the ratio of their miles.

Then $\frac{a / 154}{b / 170}=1.104$

$ \Rightarrow \frac{a}{b}=1$

Thus $50 \%$ by weight $CCl _{4}$ liquid is present in mixture.