Question

If the elevation in boiling point of a solution of $10$ g of solute(mol. wt.$=100$) in $100$ g of water is $\Delta T_b$, the ebullioscopic constant of water is:

• A. $\frac{\Delta T_b}{10}$
• B. $\Delta T_b$
• C. $10\Delta T_b$
• D. $100\Delta T_b$

Answer 1

The correct option is B $\Delta T_b$

If the elevation in boiling point of a solution of $10$ g of solute(mol. wt.$=100$) in $100$ g of water is $\Delta T_b$, the ebullioscopic constant of water is $\Delta T_b$.

The expression for the elevation in boiling point is $\Delta T_b=K_{b}m$

$K_b$ is the ebullioscopic constant of water and m is the molality of the solution.

$m=\frac{\text{Mass of solute}}{\text{molar mass of solute}\times \text{mass of solvent in kg}}=\frac{10}{100\times \frac{100}{1000}}=1m$

Hence, $K_b=\Delta T_b$

So, the correct option is $B$