What is the mole fraction of benzene in solution containing 20% by mass in $CCl_4$?

$\begin{array}{1 1}(a)\;0.66\\(b)\;0.22\\(c)\;0.33\\(d)\;0.11\end{array}$

Mole fraction of a component = $\large\frac{\text{Number of moles of the component}}{\text{ Total number of moles of all the components}}$
$\Rightarrow$ Mole fraction of benzene $=\large\frac{n_{C_6H_6}}{n_{C_6H_6}+n_{CCl_4}}$
If we start with a total mass of 100g, the mass of Benzene will be 20% of 100 = 20g, and thus the mass of $CCl_4$ will be 100 - 20 = 80g.
Molar mass of Benzene $C_6H_6 = 6 \times 12 + 6 \times 1 = 78 \;g\;mol^{-1}$
$\Rightarrow n_{C_6H_6} = \large\frac{20}{78} $$=0.2564 Molar mass of CCl_4 = 1\times 12 + 4 \times 35.4 = 154\;g \;mol^{-1} \Rightarrow n_{CCl_4} = \large\frac{80}{154}$$ =0.5194$
$\Rightarrow$ Mole fraction of benzene $= \large\frac{0.2564}{0.2564+0.5194}$$= 0.33$
edited Jul 15, 2014