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Question

When 20 g of naphtholic acid (C11H8O2) is dissolved in 50 g of benzene (Kf = 1.72 K kg mol−1), a freezing point depression of 2 K is observed. The van’t Hoff factor (i) is

Options

A.0.5
B.1
C.2
D.3

Solution

$\Delta T_{f\left( \text{theo} \right)} = K_{f}.m = 1.72 \times \frac{20/172}{50/1000} = 4\text{ K}$

Now, $i = \frac{\Delta T_{f\left( \text{exp} \right)}}{\Delta T_{f\left( \text{theo} \right)}} = \frac{2}{4} = 0.5$

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