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The freezing point depression constant for water is $-1.86^{\circ} \mathrm{C} \mathrm{m}^{-1}$. If $5.00 \mathrm{~g} \mathrm{Na} \mathrm{SO}_4$ is dissolved in $45.0 \mathrm{~g} \mathrm{H}_2 \mathrm{O}$, the freezing point is changed by $-3.82^{\circ} \mathrm{C}$. Calculate the van't Hoff factor for $\mathrm{Na}_2 \mathrm{SO}_4$.
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Verified Answer
The correct answer is:
2.63
$$
\begin{aligned}
\Delta T_f & =i \times K_f \cdot m \\
\text { or } i & =\frac{\Delta T_f \times W_{\text {solvent }}}{K_f \times n_{\text {solute }} \times 1000} \\
& =\frac{3.82 \times 45}{1.86 \times\left(\frac{5}{142}\right) \times 1000} \\
& =2.63
\end{aligned}
$$
\begin{aligned}
\Delta T_f & =i \times K_f \cdot m \\
\text { or } i & =\frac{\Delta T_f \times W_{\text {solvent }}}{K_f \times n_{\text {solute }} \times 1000} \\
& =\frac{3.82 \times 45}{1.86 \times\left(\frac{5}{142}\right) \times 1000} \\
& =2.63
\end{aligned}
$$
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