Question

The enthalpy of vapourisation of water is $9700calmo{l}^{-1}$. Given that pure water boils at ${100}^{\circ }C$, then calculate the value of ebullioscopic constant of water :

• A. $0.51Kkgmo{l}^{-1}$
• B. $1.21Kkgmo{l}^{-1}$
• C. $0.15Kkgmo{l}^{-1}$
• D. $2.01Kkgmo{l}^{-1}$

Answer 1

The correct option is A $0.51Kkgmo{l}^{-1}$

$K_b=\frac{R\times T_b^2\times M}{1000\times \Delta H_{vap}}$
where,
$T_b\text{is boiling point of pure solvent}={100}^{\circ }C=373K$
$M\text{is molar mass of water}=18gmo{l}^{-1}$
$\Delta H_{vap}\text{is the molar enthalpy of vapourisation of water}=9700calmo{l}^{-1}$
$R\text{is ideal gas constant}=1.987calmo{l}^{-1}{K}^{-1}$
$K_b=\frac{1.987\times (373{)}^2\times 18}{1000\times 9700}{K}_b=0.513Kkgmo{l}^{-1}$