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 $\mathrm{H}_2 \mathrm{O}$ $$ \begin{aligned} \mathrm{X}_{\text {solute }} & =\frac{1}{\frac{1000}{18}+1}=\frac{18}{1018} \\ & =0.01768 \approx 0.018 . \end{aligned} $$ Related Theory The molar fraction can be represented by $X$. 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'.