Question

CHEMISTRY
Q1 The freezing point of benzene decreases by 2.12 K when 2.5 g of benzoic acid (C6H5COOH) is
dissolved in 25 g of benzene. If benzoic acid forms a dimer in benzene, calculate the van't Hoff factor
and the percentage association of benzoic acid. (Kf for benzene = 5.12 K kg moll)

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Answer 1

Kf=5.12,deltaTf=2.12,w1=25,w2=2.5,

M2=?

We knew the equation

M2= (kf×w2×1000)/(deltaTf ×w1)

M2= (5.12×2.5×1000)/(25×2.12)

By solving we get

M2=241.50

i (van’t Hoff factor)=original molecular mass ÷ calculated molecular mass

We knew that M1 (C6H5COOH)= 122 gm per mol.

Hence i= (M1/M2) = 122/241.50

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