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$18 \mathrm{~g}$ of glucose is dissolved in $90 \mathrm{~g}$ of water. The relative lowering of vapour pressure of the solution is equal to
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The correct answer is:
$0.02$
From Rault’s law for non-volatile solute, we know that
$\frac{p^{\circ}-p_s}{p^{\circ}}=\frac{n_2}{n_1+n_2}$
For dilute solution, relative lowering of vapour pressure,
$\begin{aligned}
& n_2=\frac{18}{180}=0.1 \\
& n_1=\frac{90}{18}=5 \\
& \frac{p^{\circ}-p_s}{p^{\circ}}=\frac{0.1}{5}=0.02
\end{aligned}$
$\frac{p^{\circ}-p_s}{p^{\circ}}=\frac{n_2}{n_1+n_2}$
For dilute solution, relative lowering of vapour pressure,
$\begin{aligned}
& n_2=\frac{18}{180}=0.1 \\
& n_1=\frac{90}{18}=5 \\
& \frac{p^{\circ}-p_s}{p^{\circ}}=\frac{0.1}{5}=0.02
\end{aligned}$
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