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The freezing point of benzene decrease by 0.45C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be: (Kf for benzene =5.12 K kg mol1)

A
64.6%
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B
80.4%
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C
74.6%
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D
94.6%
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Solution

The correct option is C 94.6%
Tf=0.45 0.2g of acetic acid in 20g of benzene.
To form dimer, 2(CH3COO)CH3COO2
i=1+(1n1)a=1+(12)a=1a2
Tf=iKfm
0.45=(1a2)(5.12)(0.260201000)
1a2=0.527
a=0.945
Percentage of association is 94.5%

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