Question

Choose the correct relation for boiling point elevation constant $\left(K_b\right)$
where :

$T_b\text{is boiling point of pure solvent}$
$M\text{is molar mass of the solvent}$
$L_{vap}\text{is the molar enthalpy of vapourisation per gram of the solvent}$
$R\text{is ideal gas constant}$
(All quantities to be taken in SI units)

• A. $K_b=\frac{R^2\times T_b\times M}{1000\times L_{vap}}$
• B. $K_b=\frac{L_{vap}\times M}{1000\times R\times T_b^2}$
• C. $K_b=\frac{R\times T_b^2}{1000\times L_{vap}}$
• D. $K_b=\frac{R\times T_b}{1000\times L_{vap}}$

Answer 1

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

$K_b$ 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.
$K_b=\frac{R\times T_b^2\times M}{1000\times \Delta H_{vap}}=\frac{R\times T_b^2}{1000\times L_{vap}}$

$\Delta H_{vap}\text{is the molar enthalpy of vapourisation of the solvent}$