IIT-JEE 2007 · Paper 1 · Chemistry Q30

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ChemistrySingle correct+3 / -1Question 30 of 66

When 20 g of naphthoic acid ( $\ce{C11H8O2}$ ) is dissolved in 50 g of benzene ( $K_f=1.72$ K kg mol⁻¹), a freezing point depression of 2 K is observed. The van't Hoff factor ( $i$ ) is

0.5

Answer and worked solution

Official answer: A

Molality. $M(\ce{C11H8O2})=172$ g mol⁻¹, so $n=\frac{20}{172}=0.1163$ mol in 0.050 kg: $m=2.326$ mol kg⁻¹.

Expected depression for $i=1$ : $K_fm=1.72\times2.326=4.0$ K.

van't Hoff factor: $i=\dfrac{2}{4}=0.5$ (the acid dimerises in benzene).

(A)

Topics:van't Hoff factor Colligative properties

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IIT-JEE 2007 · Paper 1, Chemistry question 30