When 20 g of naphtholic acid (C11H8O2) is dissolved in 50 g of benzene (Kf = 1.72 K kg mol−1), a fre

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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

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$\Delta T_{f\left( 	ext{theo} ight)} = K_{f}.m = 1.72 	imes \frac{20/172}{50/1000} = 4	ext{ K}$Now, $i = \frac{\Delta T_{f\left( 	ext{exp} ight)}}{\Delta T_{f\left( 	ext{theo} ight)}} = \frac{2}{4} = 0.5$

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