The molality of an aqueous solution of methanol (CH₃OH) is 1.44 mol kg^-1. The mole fraction of methanol in the solution will be

Let the mass of solvent be l kg. We will have n₂ = 1.44mol and m₁ = 1000g

The amount of water will be $n_1=\frac{m_1}{M_1}=\frac{1000\mathrm{g}}{18\mathrm{g}{\mathrm{mol}}^{-1}}=55.56\mathrm{mol}$ 

The mole fraction of the solute will be $x_2=\frac{n_2}{n_1+n_2}=\frac{1.44\mathrm{mol}}{(55.56+1.44)\mathrm{mol}}=0.0253$