$3.1 \mathrm{~g}$ of a compound, ' $X$ ' (molar mass $=62 \mathrm{~g}$ $\mathrm{mol}^{-1}$ ) is dissolved in…
$3.1 \mathrm{~g}$ of a compound, ' $X$ ' (molar mass $=62 \mathrm{~g}$ $\mathrm{mol}^{-1}$ ) is dissolved in $19.5 \mathrm{~g}$ of other compound, $Y$ (molar mass $=78 \mathrm{~g} \mathrm{~mol}^{-1}$ ). The ratio of mole fractions of $X$ and $Y$ in the solution is
$1: 5$
$5: 1$
$4: 1$
$1: 4$
Solution
( ) The number of moles of
$
X=\frac{W_X}{M_X}=\frac{3.1 \mathrm{~g}}{62 \mathrm{~g} \mathrm{~mol}^{-1}}=0.05 \mathrm{~mol}
$
The number of moles of
$
Y=\frac{W_Y}{M_Y}=\frac{19.5 \mathrm{~g}}{78 \mathrm{~g} \mathrm{~mol}^{-1}}=0.25 \mathrm{~mol}
$
Mole fraction of $X=\frac{\text { Moles of } X}{\text { Total moles }}=\frac{0.05}{0.75}=0.066$
Mole fraction of $Y=\frac{\text { Moles of } Y}{\text { Total moles }}=\frac{0.25}{0.75}=0.33$
The ratio of mole fraction of $X$ and $Y=\frac{0.066}{0.33}=\frac{1}{5}$