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The vapour pressure of solvent decreases by $10 \mathrm{~mm} \mathrm{Hg}$ if mole fraction of non volatile solute is $0.2$ Calculate vapour pressure of solvent.
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Verified Answer
The correct answer is:
$50 \mathrm{~mm}$ of $\mathrm{Hg}$
(A)
$\begin{array}{l}
\mathrm{P}_{0}-\mathrm{P}=10 \mathrm{~mm} \text { of } \mathrm{Hg}, \quad \mathrm{x}_{2}=0.2 \\
\mathrm{P}_{0}=?
\end{array}$
According to Raoult's Law
$\begin{array}{l}
\frac{\mathrm{P}_{0}-\mathrm{P}}{\mathrm{P}_{0}}=\mathrm{x}_{2} \quad \therefore \frac{10 \mathrm{~mm} \text { of } \mathrm{Hg}}{\mathrm{P}_{0}}=0.2 \\
\therefore \mathrm{P}_{0}=50 \mathrm{~mm} \text { of } \mathrm{Hg}
\end{array}$
$\begin{array}{l}
\mathrm{P}_{0}-\mathrm{P}=10 \mathrm{~mm} \text { of } \mathrm{Hg}, \quad \mathrm{x}_{2}=0.2 \\
\mathrm{P}_{0}=?
\end{array}$
According to Raoult's Law
$\begin{array}{l}
\frac{\mathrm{P}_{0}-\mathrm{P}}{\mathrm{P}_{0}}=\mathrm{x}_{2} \quad \therefore \frac{10 \mathrm{~mm} \text { of } \mathrm{Hg}}{\mathrm{P}_{0}}=0.2 \\
\therefore \mathrm{P}_{0}=50 \mathrm{~mm} \text { of } \mathrm{Hg}
\end{array}$
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