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Calculate van't Hoff factor of $\mathrm{K}_2 \mathrm{SO}_4$ if $0.1 \mathrm{~m}$ aqueous solution of $\mathrm{K}_2 \mathrm{SO}_4$ freezes at $-0.43{ }^{\circ} \mathrm{C}$ and cryoscopic constant of water is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$.
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Verified Answer
The correct answer is:
2.3
$$
\begin{array}{ll}
& \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{K} \mathrm{K}_{\mathrm{f}} \mathrm{m} \\
\therefore \quad & 0.43=\mathrm{i} \times 1.86 \times 0.1 \\
\therefore \quad & \mathrm{i}=\frac{0.43}{1.86 \times 0.1}=2.3
\end{array}
$$
[Note: In the question, the freezing point of aqueous solution is changed from $-0.43 \mathrm{~K}$ to $-0.43^{\circ} \mathrm{C}$ to apply appropriate textual concepts.]
\begin{array}{ll}
& \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{K} \mathrm{K}_{\mathrm{f}} \mathrm{m} \\
\therefore \quad & 0.43=\mathrm{i} \times 1.86 \times 0.1 \\
\therefore \quad & \mathrm{i}=\frac{0.43}{1.86 \times 0.1}=2.3
\end{array}
$$
[Note: In the question, the freezing point of aqueous solution is changed from $-0.43 \mathrm{~K}$ to $-0.43^{\circ} \mathrm{C}$ to apply appropriate textual concepts.]
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