Answer
A - Methanol ($CH_{3}OH$)
B - Methyl Chloride ($CH_{3}Cl$)
C - Propane ($C_{3}H_{8}$)
Work Step by Step
From the Clausius-Clapeyron equation,
$ln P = -\frac{\Delta H_{vap}}{RT} + C$
The slope is $-\frac{\Delta H_{vap}}{RT}$. The steeper the slope, the greater the heat of vaporization.
Methanol has hydrogen bonding, methyl chloride has dipole-dipole interaction, and propane has dispersion forces. Methanol has the strongest intermolecular forces, followed by methyl chloride and propane. Thus, methanol has highest $\Delta H_{vap}$ and steeper slope. Propane has least $\Delta H_{vap}$ and thus has the least steep slope.
Therefore,
A - Methanol ($CH_{3}OH$)
B - Methyl Chloride ($CH_{3}Cl$)
C - Propane ($C_{3}H_{8}$)
