Question

If a pure solvent A boils at ${127}^{\circ }C$, then calculate the value of ebullioscopic constant of that solvent.
Given :The enthalpy of vapourisation of solvent is $8000calmo{l}^{-1}$
Molar mass of the solvent is $80g/mol$
$R\text{(ideal gas constant)}\approx 2calmo{l}^{-1}{K}^{-1}$

• A. $0.8Kkgmo{l}^{-1}$
• B. $1.6Kkgmo{l}^{-1}$
• C. $2.4Kkgmo{l}^{-1}$
• D. $3.2Kkgmo{l}^{-1}$

Answer 1

The correct option is D $3.2Kkgmo{l}^{-1}$

$K_b=\frac{R\times T_b^2\times M}{1000\times \Delta H_{vap}}$
where,
$T_b\text{is boiling point of pure solvent}={127}^{\circ }C=400K$
$M\text{is molar mass of solvent}=80gmo{l}^{-1}$
$\Delta H_{vap}\text{is the molar enthalpy of vapourisation of solvent}=8000calmo{l}^{-1}$
$R\text{is ideal gas constant}=2calmo{l}^{-1}{K}^{-1}$
$K_b=\frac{2\times (400{)}^2\times 80}{1000\times 8000}{K}_b=3.2Kkgmo{l}^{-1}$