Question

The values of $K_b$ and $K_f$ for solvents obeying Trouton’s rule are directly proportional to:
(A) ${\displaystyle \frac{RT}{M}}$
(B) $MT$
(C) $MRT$
(D) $M{T}^2$

Answer 1

Hint: In the colligative properties of the solvents, $K_b$ is the Ebullioscopic constant or the boiling point elevation constant while $K_f$ is the Cryoscopic Constant or the freezing point depression constant.

Formula Used: $K_b={\displaystyle \frac{R{T^2}_{boiling}{M_s}_{olvent}}{1000\mathrm{\Delta }{H}_{vap}}}$ and $K_f={\displaystyle \frac{R{T^2}_{freezing}{M_s}_{olvent}}{1000\mathrm{\Delta }{H}_{fus}}}$
Where, R is the universal gas constant, $T_b$ is the boiling point of the liquid, $M_{solvent}$ is the molecular weight of the solvent, $T_f$ is the freezing point of the liquid, $\mathrm{\Delta }{H}_{vap}$ and $\mathrm{\Delta }{H}_{fus}$ are the heats of vaporization and fusion, respectively.

Complete step by step answer:
The Trouton’s rule states that for many (but not all) the liquids, the entropy of vaporization is approximately the same at $85$ $J^{-1}mo{l}^{-1}{K}^{-1}$ and the success of this rule lies in the fact that the entropy of a gas is considerably larger than that of a liquid.
Mathematically, $S_{gas}$ >> $S_{liquid}$ .
Therefore the entropy of a liquid is negligible in determining the entropy of vaporization.
 $\mathrm{\Delta }{S}_{vap}=S_{gas}-S_{liquid}\approx S_{gas}$
When a liquid vaporizes, its entropy changes drastically and is related to the enthalpy of vaporization and the transition temperature as,
 $\mathrm{\Delta }{S}_{vap}$ = ${\displaystyle \frac{\mathrm{\Delta }{H}_{vap}}{T}}$ $\approx$ $85$ $J^{-1}mo{l}^{-1}{K}^{-1}$ which is a constant value.
Mathematically,
 $K_b={\displaystyle \frac{R{T^2}_{boiling}{M_s}_{olvent}}{1000\mathrm{\Delta }{H}_{vap}}}$ while $K_f={\displaystyle \frac{R{T^2}_{freezing}{M_s}_{olvent}}{1000\mathrm{\Delta }{H}_{fus}}}$ .
Hence for solvents or liquids following Trouton’s Rule we get, putting the value of ${\displaystyle \frac{\mathrm{\Delta }{H}_{vap}}{T}}$ as a constant in the above equations,
 $K_b\approx M{T}_b$ and all the other quantities were taken as constants.
Similarly for $K_f=M{T}_f$ .
Hence, the correct answer is option B.

Note:
Trouton’ Rule is not applicable to liquids such as acetic acid, ethanol, water, formic, acid, etc. which undergo association and dissociation in water through hydrogen bonding because the enthalpy of vaporization is more for hydrogen-bonding liquids and for low molecular weight solvents this effect is more pronounced.