Question

The correct relation between elevation of boiling point and molar mass of solute is

• A. $M_{2} = \frac{K_{b} . W_{2}}{\Delta T_{b}.W_{1}} $
• B. $M_{2} = \frac{K_{b} . W_{1}}{\Delta T_{b}.W_{2}} $
• C. $M_{2} = \frac{T_{b} . K_{b}}{W_{1}.W_{2}} $
• D. $M_{2} = \frac{\Delta T_{b} . W_{1}}{K_{b}.W_{2}} $

Answer 1

Answer: (A)

For dilute solutions, the elevation in boiling point is directly proportional to the molal concentration of the solute in solution $(m)$

Thus, $\Delta T_{b} \propto m $

$\Delta T_{b}=k_{b} m$

Where, $k_{b}$ is molal boiling point elevation constant.

As, molality $(m)=\frac{\text { Moles of solute }\left(n_{2}\right)}{\text { Mass of solvent }\left(W_{1}\right)(\text { in } kg )}$

Moles of solute $\left(n_{2}\right)=\frac{W_{2}}{M_{2}}$

$\therefore \Delta T_{b}=\frac{k_{b} \times W_{2}}{M_{2} \times W_{1}}$

$\Rightarrow M_{2}=\frac{k_{b} \cdot W_{2}}{\Delta T_{b} \cdot W_{1}}$