IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

shinee

When 20g of Naphthoic acid (C₁₁ Hg O₂ ) iis dissolved in 50g of benzene (k_f +1.72KKgmol^-1 ), a freezing point depressionof 2K is observed .the van't hoff factor is
a)0.5
b)2.0
c)1.0
d)3.0

LAMPARD

Getting it as 1.06...is it c??

shinee

i dont know the answer, can u post the solution plz

viswanath

im gettin around 1.059. so, ans shud be c.

Solution: $\Delta$ T_f = i * K_f* m, where m is the molality and i is the Vant-Hoff Factor

First lets find molality, m = (20 * 1000) $\div$ (50 * 364) = 1.098 (364 is the molecular wt of Naphthoic acid)

so, i = $\Delta$ T_f/ (K_f* m). substituting the values, m = 1.098, K_f= 1.72, $\Delta$ T_f = 2 gives i as 1.059

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