Question

The mole fraction of $CC{l}_4(g)$ in the vapour in equilibrium with liquid mixture of $CC{l}_4$ and $SiC{l}_4$ is $\mathrm{0.3474.}$ The vapour pressure of $SiC{l}_4$ and $CC{l}_4$ is $238.3$ and $114.9mm$, respectively, at the same temperature. Calculate $\%$ by weight of $CC{l}_4$ in liquid mixture.

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

Answer 1

Answer: (A)

$X_{CC{l}_4}(g)=0.3474$

$X_{SiC{l}_4}(g)=0.6526$

$P_{SiC{l}_4(g)}^{\circ }=238.3mm$

$P_{CC{l}_4}^{\circ }=114.9mm$

$P_{CC{l}_4}^{\prime }=P_{CC{l}_4}^{\circ }\cdot X_{CC{l}_4(\ell )}=P_M\times X_{CC{l}_4(g)}$

$\therefore 144.9\times X_{CC{l}_4(\ell )}=P_M\times 0.3474\dots$ (i)

Also, $238.3\times X_{SiC{l}_4(\ell )}=P_M\times 0.6526\dots$(ii)

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

$\frac{X_{CC{l}_4(\ell )}}{X_{SiC{l}_4(\ell )}}$

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

Let $agCC{l}_4$ and $bgSiC{l}_4$ 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 $CC{l}_4$ liquid is present in mixture.