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Question

What is the value of van’t Hoff factor (i) for 60% association of CH3COOH in benzene?


A
0.5
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B
0.4
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C
0.7
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D
0.9
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Solution

The correct option is C 0.7
In benzene, two molecules of ethanoic acid associate to form a dimer.
2CH3COOH(CH3COOH)2
Initial     a                  0At.eq      a(1α)     aα2
Number of moles at eq. neq=a(1α+α/2)
At 60 % association α=0.6
neq=a(10.6+0.3)=a(0.7)

van't Hoff factor i = number of solute particles in solutiontheoritical number of solute particles
 i=a(0.7)a=0.7 

Chemistry

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