The mole fraction of methanol (CH₃OH) in aqueous solution is 0.10. If density of the solution is 0.97 g mL^-1 , the molarity of the solution will be

For x₂ = 0.10, let we have n₂ = 0.1 mol and n₁ = 0.9 mol

The mass of solution will be

$m=n_1{M}_1+n_2{M}_2=\left(0.9\mathrm{mol}\right)\left(18\mathrm{g}{\mathrm{mol}}^{-1}\right)+\left(0.1\mathrm{mol}\right)\left(32{\mathrm{mol}}^{-1}\right)=19.4\mathrm{g}$

The volume of solution will be

$V=\frac{m}{\rho }=\frac{19.4\mathrm{g}}{0.97{\mathrm{gmL}}^{-1}}=20.0\mathrm{ml}=20\times {10}^{-3}\mathrm{L}$

The molarity of solution will be

$M=\frac{n_2}{V}=\frac{0.1\mathrm{mol}}{20\times {10}^{-3}\mathrm{L}}=5.0{\mathrm{molL}}^{-1}$