Question

A solution is prepaired by mixing $8.5g$ of $C{H}_{2}C{l}_2$ and $11.95g$ of $CHC{l}_3$. If vapour pressure of $C{H}_{2}C{l}_2$ and $CHC{l}_3$ at $4298K$ are $415$ and $200mmHg$ respectively, the mole fraction of $CHC{l}_3$ in vapour form is : (molar mass of $Cl=35.5gmo{l}^{-1}$)

• A. $0.162$
• B. $0.675$
• C. $0.325$
• D. $0.486$

Answer 1

The correct option is C $0.325$

Molar mass of $C{H}_{2}C{l}_2$, $M_1=85g(mol{)}^{-1}$

moles of $C{H}_{2}C{l}_2\left(n_1\right)$=$\frac{W_1}{M_1}=\frac{8.5g}{85g(mol{)}^{-1}}=0.1mol$

Molar mass of $CHC{L}_3,M_2=119.5g(mol{)}^{-1}$

Moles of $CHC{l}_3\left(n_2\right)=\frac{W_2}{M_2}=\frac{11.95g}{119.5g(mol{)}^{-1}}=0.1mol$

Mole fraction of $CHC{l}_3\left(x_2\right)$=$\frac{n_2}{n_1+n_2}=\frac{0.1mol}{0.2mol}=0.5$

Partial vapour pressure of $CHC{l}_3\left(p_2\right)=x_2{p}_2^0=0.5\times 200mmHg=100mmHg$

Partial pressure of $CHC{l}_2\left(p_1\right)=(1-x_2)p_1^0=0.5\times 415mmHg=207.5mmHg$

Mole fraction of $CHC{l}_3$ in vapour phase is $y_2=\frac{p_2}{p_1+p_2}=\frac{100mmHg}{307.5mmHg}=0.325$