The mole friction of the solute in one molal aqueous solution is:
The mole friction of the solute in one molal aqueous solution is:
0.009
0.018
0.027
0.036
Solution
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'.