Question

A solution is prepared by mixing 8.5 g of $C{H}_{2}C{l}_2$ and 11.95 g of $CHC{l}_3$. If vapour pressure of $C{H}_{2}C{l}_2$ and $CHC{l}_3$ at 298 K are 415 and 200 mmHg respectively, the mole fraction of $CHC{l}_3$ in vapour form is:

[Molar mass of C1=35.5 g $mo{l}^{-1}$]

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

Answer 1

The correct option is A 0.325

Given

$P_0=$ Vapor pressure = 200 mm of Hg

Mass of solute (CH${}_2$Cl${}_2$)=8.5 g

Mass of solvent (CHCl${}_3$)=11.95 g

Mole fraction in solution phase can be calculated as:

$X_{C{H}_{2}C{l}_2}=\frac{\left(8.5/85\right)}{\left(8.5/85\right)+\left(11.95/119.5\right)}=0.5$

$X_{CHC{l}_3}=\frac{\left(11.95/119.5\right)}{\left(8.5/85\right)+\left(11.95/119.5\right)}=0.5$

Mole fraction in vapor phase is given by the formula:

$Y_{CHC{l}_3}=\frac{P^o.X_{CHC{l}_3}}{P^o.X_{CHC{l}_3}+P^o.X_{C{H}_{2}C{l}_2}}=\frac{200\times 0.5}{200\times 0.5+415\times 0.5}=0.325$

Hence, the correct option is $A$