Question

Calculate the value of ebullioscopic constant of a solvent if enthalpy of vapourisation of that solvent is nearly $3974calmo{l}^{-1}$.
Given :
Boiling point of pure solvent is ${127}^{\circ }C$
Molecular mass of the solvent is $20gmo{l}^{-1}$

• A. $1.6Kkgmo{l}^{-1}$
• B. $0.5Kkgmo{l}^{-1}$
• C. $2.4Kkgmo{l}^{-1}$
• D. $3.2Kkgmo{l}^{-1}$

Answer 1

The correct option is A $1.6Kkgmo{l}^{-1}$

$K_b=\frac{R\times T_b^2\times M}{1000\times \Delta H_{vap}}$
where,
$T_b\text{is boiling point of pure solvent}={127}^{\circ }C=400K$
$M\text{is molar mass of the solvent}=20gmo{l}^{-1}$
$\Delta H_{vap}\text{is the molar enthalpy of vapourisation of solvent}=3974calmo{l}^{-1}$
$R\text{is ideal gas constant}=1.987calmo{l}^{-1}{K}^{-1}$

$K_b=\frac{1.987\times (400{)}^2\times 20}{1000\times 3974}{K}_b=1.6Kkgmo{l}^{-1}$