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The vapour pressure of water at $20^{\circ} \mathrm{C}$ is $17.5 \mathrm{~mm} \mathrm{Hg}$. If $18 \mathrm{~g}$ of glucose $\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)$ is added to $178.2 \mathrm{~g}$ of water at $20^{\circ} \mathrm{C}$, the vapour pressure of the resulting solution will be
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Verified Answer
The correct answer is:
$17.325 \mathrm{~mm} \mathrm{Hg}$
$17.325 \mathrm{~mm} \mathrm{Hg}$
$$
\begin{aligned}
& \frac{P^0-P_s}{P_s}=X_{\text {solute }} \\
& \frac{17.5-P_s}{P_s}=\frac{0.1}{10} \\
& \frac{17.5-P_s}{P_s}=0.01 \\
& \Rightarrow P_s=17.325 \mathrm{~mm} \mathrm{Hg}
\end{aligned}
$$
\begin{aligned}
& \frac{P^0-P_s}{P_s}=X_{\text {solute }} \\
& \frac{17.5-P_s}{P_s}=\frac{0.1}{10} \\
& \frac{17.5-P_s}{P_s}=0.01 \\
& \Rightarrow P_s=17.325 \mathrm{~mm} \mathrm{Hg}
\end{aligned}
$$
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