JEE MAIN - Chemistry (2022 - 26th July Evening Shift - No. 15)
The elevation in boiling point for 1 molal solution of non-volatile solute A is $$3 \mathrm{~K}$$. The depression in freezing point for 2 molal solution of $$\mathrm{A}$$ in the same solvent is 6 $$K$$. The ratio of $$K_{b}$$ and $$K_{f}$$ i.e., $$K_{b} / K_{f}$$ is $$1: X$$. The value of $$X$$ is [nearest integer]
Answer
1
Explanation
Molality of a solution of non volatile solute $(A)=1$
Elevation in boiling point is given by
$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \mathrm{m}$
$$ 3=\mathrm{K}_{\mathrm{b}} \times 1 $$ ... (1)
Molality of $(A)$ in the same solvent $=2$
Depression in freezing point is given by
$$ \begin{aligned} &\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \mathrm{m} \\ &6=\mathrm{K}_{\mathrm{f}} \times 2 \quad\quad\quad{...(2)} \\ &\text { Dividing (1) by (2) } \\ &\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{K}_{\mathrm{f}}}=\frac{1}{\mathrm{X}}=\frac{1}{1} \\ &\therefore \quad \mathrm{X}=1 \end{aligned} $$
Elevation in boiling point is given by
$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \mathrm{m}$
$$ 3=\mathrm{K}_{\mathrm{b}} \times 1 $$ ... (1)
Molality of $(A)$ in the same solvent $=2$
Depression in freezing point is given by
$$ \begin{aligned} &\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \mathrm{m} \\ &6=\mathrm{K}_{\mathrm{f}} \times 2 \quad\quad\quad{...(2)} \\ &\text { Dividing (1) by (2) } \\ &\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{K}_{\mathrm{f}}}=\frac{1}{\mathrm{X}}=\frac{1}{1} \\ &\therefore \quad \mathrm{X}=1 \end{aligned} $$
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