Question

At $323 \mathrm{~K}$, the vapour pressure (in $\mathrm{mm}$ $\mathrm{Hg}$ ) of a methanol-ethanol solution is represented as: $$ P=120 X+140 $$ where $X$ is the mole fraction of methanol in liquid solution at equilibrium. Then the value of $P_{\mathrm{E} \text { ton }}^{\circ}$ is (a) $120 \mathrm{~mm}$ (b) $140 \mathrm{~mm}$ (c) $260 \mathrm{~mm}$ (d) $20 \mathrm{~mm}$

          At $323 \mathrm{~K}$, the vapour pressure (in $\mathrm{mm}$ $\mathrm{Hg}$ ) of a methanol-ethanol solution is represented as:
$$
P=120 X+140
$$
where $X$ is the mole fraction of methanol in liquid solution at equilibrium. Then the value of $P_{\mathrm{E} \text { ton }}^{\circ}$ is
(a) $120 \mathrm{~mm}$
(b) $140 \mathrm{~mm}$
(c) $260 \mathrm{~mm}$
(d) $20 \mathrm{~mm}$

Added by Zachary P.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Lijeesh Krishnan

Step 1

We know that the mole fraction of methanol (X) and ethanol (Y) in the liquid solution must add up to 1, i.e., $X + Y = 1$. Show more…

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At $323 \mathrm{~K}$, the vapour pressure (in $\mathrm{mm}$ $\mathrm{Hg}$ ) of a methanol-ethanol solution is represented as: $$ P=120 X+140 $$ where $X$ is the mole fraction of methanol in liquid solution at equilibrium. Then the value of $P_{\mathrm{E} \text { ton }}^{\circ}$ is (a) $120 \mathrm{~mm}$ (b) $140 \mathrm{~mm}$ (c) $260 \mathrm{~mm}$ (d) $20 \mathrm{~mm}$