Question

Which of the following relation is correct for $\text{boiling point elevation constant}\left(K_b\right)$?
Here
$T_b\text{is boiling point of pure solvent}$
$M\text{is molar mass of the solvent}$
$\Delta H_{vap}\text{is the molar enthalpy of vapourisation of the solvent}$
$R\text{is ideal gas constant}$
(All quantities to be taken in SI units)

• A. $K_b=\frac{R^2\times T_b\times M}{1000\times \Delta H_{vap}}$
• B. $K_b=\frac{\Delta H_{vap}\times M}{1000\times R\times T_b^2}$
• C. $K_b=\frac{R\times T_b^2\times M}{1000\times \Delta H_{vap}}$
• D. None of the above

Answer 1

The correct option is C $K_b=\frac{R\times T_b^2\times M}{1000\times \Delta H_{vap}}$

Theory:
At B.P, the vapour pressure of the liquid becomes equal to the atmospheric pressure
Let a non-volatile solute B be added to the pure liquid A.
After addition of B, the vapour pressure of the pure solvent (A) must decrease.
On addition of a non-volatile substance to a pure liquid, the B.P of the solution becomes higher than that of the pure liquid. So, the B.P. is elevated.

If:
$T_b\text{is boiling point of solution}$
$T_b^{\circ }\text{is boiling point of pure liquid}$
then ,
Elevation in Boiling Point
$\Delta T_b=T_b-T_b^{\circ }$
Vapour Pressure (VP) of solution is less than that of pure solvent.
Temperature, at which VP equals atmospheric pressure, is higher for solution. Boiling point of the solution is higher by an amount $\Delta T_b$.

Elevation of BP is only dependent on the solute concentration and not its nature.
From experimental observations :
$\Delta T_b=K_b\times m$
Here $m\text{is molality}$
$K_b$ is Boiling Point Elevation Constant or Molal Elevation Constant
It is also known as ebullioscopic constant.
Unit of Ebullioscopic constant is $Kkgmo{l}^{-1}$.
It is the property of the solvent and is independent of the solute.
$K_b=\frac{R\times T_b^2\times M}{1000\times \Delta H_{vap}}=\frac{R\times T_b^2}{1000\times L_{vap}}$
$L_{vap}\text{is the latent heat of vapourisation of the solvent}$