A) 64.6%
B) 80.4%
C) 74.6%
D) 94.6%
Correct Answer: D
Solution :
In benzene \[2C{{H}_{3}}COOH\rightleftharpoons {{(C{{H}_{3}}COOH)}_{2}}\] \[i=1+\left( \frac{1}{2}-1 \right)\alpha \] \[i=1-\frac{\alpha }{2}\]Here \[\alpha \]is degree of association \[\Delta {{T}_{f}}=i{{K}_{f}}m\]\[0.45=\left( 1-\frac{\alpha }{2} \right)(5.12)\frac{\left( \frac{0.2}{60} \right)}{\frac{20}{1000}}\] \[1-\frac{\alpha }{2}=0.527\] \[\alpha =0.945\] % degree of association =94.5%You need to login to perform this action.
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