Question

$1g$ of non-volatile non-electrolyte solute is dissolved in $100g$ of two different solvents $A\text{and}B$ whose ebullioscopic constants are in the ratio of $1:5$ . The ratio of the elevation in their boiling point, $\frac{\Delta T_{b} \left(A\right)}{\Delta T_{b} \left(B\right)},$ is

• A. $5:1$
• B. $10:1$
• C. $1:5$
• D. $1:3$

Answer 1

Answer: (C)

$\Delta T_{b}=K_{b}\times m\Rightarrow \Delta T_{b} \propto K_{b}$
As $m_{A}=m_{B}$ (given)
So $\frac{\Delta T_{b \left(A\right)}}{\Delta T_{b \left(B\right)}}=\frac{K_{b \left(A\right)}}{K_{b \left(B\right)}}$
As $\frac{K_{b \left(A\right)}}{K_{b \left(B\right)}}=\frac{1}{5}$
So $\frac{\Delta T_{b \left(A\right)}}{\Delta T_{b \left(B\right)}}=\frac{1}{5}$