Question

The boiling point of pure solvent is is ${27}^{\circ }C$ and its heat of vapourisation is $84cal/g$. Determine the boiling point elevation constant $\left(K_b\right)$ for the solvent

• A. $6.89Kkgmo{l}^{-1}$
• B. $0.23Kkgmo{l}^{-1}$
• C. $2.14Kkgmo{l}^{-1}$
• D. $5.18Kkgmo{l}^{-1}$

Answer 1

The correct option is C $2.14Kkgmo{l}^{-1}$

We know,
Boiling point elevation constant,
$K_b=\frac{R{T}_b^2}{1000L_v}$
where,
$T_b$ is boiling point of pure solvent in $K$
$R\text{in}cal/(mol.K)$
$L_v$ is latent heat of vaourisation of solvent in $cal/g$
Thus,
$K_b=\frac{2\times (27+273{)}^2}{1000\times 84}=2.14Kkgmo{l}^{-1}$