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$3.1 \mathrm{~g}$ 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
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The correct answer is:
$1: 5$
( ) 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 $Y=\frac{0.066}{0.33}=\frac{1}{5}$.
$$
X=\frac{W_X}{M_X}=\frac{3.1 \mathrm{~g}}{62 \mathrm{~g} \mathrm{~mol}^{-1}}=0.05 \mathrm{~mol}
$$
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 $Y=\frac{0.066}{0.33}=\frac{1}{5}$.
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