A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of $290 \mathrm{~mm}$ at $300…
A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of $290 \mathrm{~mm}$ at $300 \mathrm{~K}$.
The vapour pressure of propyl alcohol is $200 \mathrm{~mm}$. If the mole fraction of ethyl alcohol is $0.6$, its vapour pressure (in $\mathrm{mm}$) at the same temperature will be
$350$
$300$
$700$
$360$
Solution
Let the vapour pressure of pure ethyl alcohol be $\mathrm{P}$,
According to Raoult's law
$\begin{aligned}
& 290=200 \times 0.4+\mathrm{P} \times 0.6 \\
& \mathrm{P}=\frac{290-80}{0.6}=350 \mathrm{~mm} \mathrm{~Hg}
\end{aligned}$
Hence, (A) is correct.