Question

$y$ gm of a non-volatile organic substance of molecular mass $M$ is dissolved in 250 gm benzene. The molal elevation constant of benzene is $K_b$. The elevation in its boiling point is given by:

• A. $\frac{M}{K_{b}y}$
• B. $\frac{4K_{b}y}{M}$
• C. $\frac{K_{b}y}{4M}$
• D. $\frac{K_{b}y}{M}$

Answer 1

The correct option is B $\frac{4K_{b}y}{M}$

$Molaliy=\frac{W_{solute}}{M\cdot wt}\times 1000W_{solv.}$
$Molality=\frac{y}{M}\times \frac{1000}{250}$
$m=\frac{4y}{M}$
$\Delta T_b=K_{b}m$$=\frac{4K_{b}y}{M}$

Hence, the correct option is $B$