JEE MAIN - Chemistry (2022 - 25th July Evening Shift - No. 2)
Two solutions A and B are prepared by dissolving 1 g of non-volatile solutes X and Y, respectively in 1 kg of water. The ratio of depression in freezing points for A and B is found to be 1 : 4. The ratio of molar masses of X and Y is
1 : 4
1 : 0.25
1 : 0.20
1 : 5
Explanation
$$\Delta T_{f}=i k_{f} \times m$$
$$ \frac{\Delta T_{\mathrm{f}(\mathrm{A})}}{\Delta \mathrm{T}_{\mathrm{f}(\mathrm{B})}}=\frac{1}{4} $$
$$\frac{\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \frac{1}{\mathrm{M}_{\mathrm{A}}} \times 1}{\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \frac{1}{\mathrm{M}_{\mathrm{B}}} \times 1}=\frac{1}{4}$$
$$\frac{M_{B}}{M_{A}}=\frac{1}{4}$$
$$M_{A}: M_{B}=4: 1$$
$$ \frac{\Delta T_{\mathrm{f}(\mathrm{A})}}{\Delta \mathrm{T}_{\mathrm{f}(\mathrm{B})}}=\frac{1}{4} $$
$$\frac{\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \frac{1}{\mathrm{M}_{\mathrm{A}}} \times 1}{\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \frac{1}{\mathrm{M}_{\mathrm{B}}} \times 1}=\frac{1}{4}$$
$$\frac{M_{B}}{M_{A}}=\frac{1}{4}$$
$$M_{A}: M_{B}=4: 1$$
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