JEE MAIN - Chemistry (2022 - 25th July Morning Shift - No. 3)
The depression in freezing point observed for a formic acid solution of concentration $$0.5 \mathrm{~mL} \mathrm{~L}^{-1}$$ is $$0.0405^{\circ} \mathrm{C}$$. Density of formic acid is $$1.05 \mathrm{~g} \mathrm{~mL}^{-1}$$. The Van't Hoff factor of the formic acid solution is nearly : (Given for water $$\mathrm{k}_{\mathrm{f}}=1.86\, \mathrm{k} \,\mathrm{kg}\,\mathrm{mol}^{-1}$$ )
0.8
1.1
1.9
2.4
Explanation
$$\Delta \mathrm{T}_{\mathrm{f}}$$ of formic acid $$=0.0405^{\circ} \mathrm{C}$$
Concentration $$=0.5 \mathrm{~mL} / \mathrm{L}$$
and density $$=1.05 \mathrm{~g} / \mathrm{mL}$$
$$\therefore$$ Mass of formic acid in solution $$=1.05 \times 0.5 \mathrm{~g}$$
$$ =0.525 \mathrm{~g} $$
$$\therefore$$ According to Van't Hoff equation,
$$ \begin{aligned} &\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{k}_{\mathrm{f}} \cdot \mathrm{m} \\ &0.0405=\mathrm{i} \times 1.86 \times \frac{0.525}{46 \times 1} \end{aligned} $$
(Assuming mass of $$1 \mathrm{~L}$$ water $$=\mathrm{kg}$$ )
$$ \mathrm{i}=\frac{0.0405 \times 46}{1.86 \times 0.525}=1.89 \approx 1.9 $$
Concentration $$=0.5 \mathrm{~mL} / \mathrm{L}$$
and density $$=1.05 \mathrm{~g} / \mathrm{mL}$$
$$\therefore$$ Mass of formic acid in solution $$=1.05 \times 0.5 \mathrm{~g}$$
$$ =0.525 \mathrm{~g} $$
$$\therefore$$ According to Van't Hoff equation,
$$ \begin{aligned} &\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{k}_{\mathrm{f}} \cdot \mathrm{m} \\ &0.0405=\mathrm{i} \times 1.86 \times \frac{0.525}{46 \times 1} \end{aligned} $$
(Assuming mass of $$1 \mathrm{~L}$$ water $$=\mathrm{kg}$$ )
$$ \mathrm{i}=\frac{0.0405 \times 46}{1.86 \times 0.525}=1.89 \approx 1.9 $$
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