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Question
Why is the value of van't Hoff factor for ethanoic acid in benzene close to 0.5?
Solution
In benzene, two molecules of ethanoic acid associate to form a dimer.
\[\ce{2CH3COOH <=> (CH3COOH2)}\]
Thus, the van't Hoff factor,
i = `"Number of solutes after dissociation or associations"/"Number of solutes before dissociation or associations"`
= `1/2`
= 0.5
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