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Calculate the relative lowering of vapour pressure if the vapour pressure of benzene and vapour pressure of solution of non-volatile solute in benzene are $640 \mathrm{mmHg}$ and $590 \mathrm{mmHg}$ respectively at same temperature.
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Verified Answer
The correct answer is:
$0.078$
Relative lowering of vapour pressure
$$
\begin{aligned}
& =\frac{\Delta \mathrm{P}}{\mathrm{P}_1^0}=\frac{\mathrm{P}_1^0-\mathrm{P}_1}{\mathrm{P}_1^0} \\
& =\frac{640-590}{640}=0.078
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{\Delta \mathrm{P}}{\mathrm{P}_1^0}=\frac{\mathrm{P}_1^0-\mathrm{P}_1}{\mathrm{P}_1^0} \\
& =\frac{640-590}{640}=0.078
\end{aligned}
$$
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