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Depression in Freezing Point

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Question

A solution containing $10$ g of a dibasic acid in $1000$ g water freezes at $272.85 K$ . $10$ mL of this solution requires $12$ mL of $N / 10$ $N a O H$ for complete neutralization. If $K_{f}$ for water is $1.86 K m o l^{- 1} k g$ . Calculate Van't Hoff factor for acid (nearest integer).

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Solution

The number of millimles of NaOH are $12 m l \times 0.1 N = 1.2 m i l l i m o l e s$
They will neutralize 0.6 millimoles of the acid.
0.6 millimoles of the acid corresponds to $\frac{10 g \times 10 m l}{1000 m l} = 0.1 g$ .
Hence, the molar mass of the acid is $\frac{0.1}{0.6} \times 1000 = 166.7 g / m o l$ .
The depression in the freezing point is $273 - 272.85 = 0.15 K$ .
The molality of the solution is $m = \frac{\frac{10 g}{166.7 g / m o l}}{1 k g} = 0.06 m o l / k g$
The expression for the depression in the freezing point is $Δ T_{f} = i K_{f} m$
Substitute values in the above expression.
$0.15 = i \times 1.86 \times 0.06$
Hence, $i = 1.344$

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