The mole friction of the solute in one molal aqueous solution is:

1 mole of solute present in $1 \mathrm{~kg}$ of $\mathrm{H}_2 \mathrm{O}$ 1 mole of solute present in $\frac{1000}{18} \mathrm{~g}$. mole of $\mathrm{H}_2 \mathrm{O}$$\begin{aligned}\mathrm{X}_{\text{solute }}& =\frac{1}{\frac{1000}{18}+1}=\frac{18}{1018}\\ & =0.01768 \approx0.018 . \end{aligned}$$Related Theory$ The molar fraction can be represented byX$. If the solution consists of components$A and B$, then the mole fraction is, Mole fraction of solute =$\frac{\text{ Moles$of$Solute }}{\text{ Moles of Solute }+ \text{ Moles of Solven }} =\frac{n_A}{n_A+n_B}$Mole fraction of solute$=\frac{n_B}{n_A+n_B} Therefore, the sum of mole fraction of all the components is always equal to one. Caution Students misread the word 'molal' as 'molar'.