The mole fraction of benzene in a solution containing $39 \%$ by mass in an organic solvent of molecular mass 122 is

Given, $39 \%$ by mass of organic solvent in benzene $\therefore 39 \mathrm{~g}$ of organic solvent in $100 \mathrm{~g}$ solution. or $39 \mathrm{~g}$ of organic solvent in $100-39=61$ g benzene. Now, molar mass of organic solvent $=122 \mathrm{~g} / \mathrm{mol}$ Molar mass of benzene $\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)=12 \times 6+6=78 \mathrm{~g} / \mathrm{mol}$ $\therefore$ Moles of benzene $\left(n_{B}\right)=\frac{61}{78}$ Moles of organic solvent $\left(n_{\mathrm{O}}\right)=\frac{39}{122}$ Mole fraction of benzene $=\frac{n_{B}}{n_{B}+n_{\mathrm{O}}}=\frac{\frac{61}{78}}{\frac{61}{78}+\frac{39}{122}}=0.6$