Assume benzene has a mass of $30 \mathrm{~g}$ in a solution with a total mass of $100 \mathrm{~g}$.

Mass of $\mathrm{CCl}_{4}=(100-30) \mathrm{g}$

$=70 \mathrm{~g}$

benzene’s molar mass $\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)=(6 \times 12+6 \times 1) \mathrm{g} \mathrm{mol}^{-1}$

$=78 \mathrm{~g} \mathrm{~mol}^{-1}$

hence, moles of $\mathrm{C}_{6} \mathrm{H}_{6}=\frac{30}{78}$ mol

$=0.3846 \mathrm{~mol}$

Molar mass of $\mathrm{CCl}_{4}=1 \times 12+4 \times 355=154 \mathrm{~g} \mathrm{~mol}^{-1}$

so, moles of $\mathrm{CCl}_{4}=\frac{70}{154}$ mol

$=0.4545 \mathrm{~mol}$

hence, the mole fraction of $\mathrm{C}_{6} \mathrm{H}_{6}$ is:

$\frac{\text { Number of moles of } C_{6} H_{6}}{\text { Number of moles of } C_{6} H_{6}+\text { Number of moles of } \mathrm{CCl}_{4}}$

$=\frac{0.3846}{0.3846+0.4545}$

$=0.458$