1. The freezing point of benzene de...
Question

# 1. The freezing point of benzene decreases by 0.45°C 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 (K, for benzene = 5.12 K kg molº 1) (2017 Main) (a) 64.6% (b) 80.4 % (c) 74.6 % (d) 94.6% D. AT

JEE/Engineering Exams
Chemistry
Solution
75
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Let the degree of association of acetic acid ( left(mathrm{CH}_{3} mathrm{COOH}right) ) in benzene is ( alpha, ) then Initial [ 2 mathrm{CH}_{3} mathrm{COOH} rightleftharpoonsleft(mathrm{CH}_{3} mathrm{COOH}right)_{2} ] Moles at equilibrium ( 1-alpha ) ( therefore ) Total moles ( =1-alpha+frac{alpha}{2}=1-frac{alpha}{2} ) or ( i=1-frac{alpha}{2} ) Now, depression in freezing point ( left(Delta T_{f}right) ) is given as ( Delta T_{f}=i K_{f} m ) where, ( K_{f}= ) molal depression constant or cryoscopic constant. ( n= ) molality Molality ( =frac{text { number of moles of solute }}{text { weight of solvent }(text { in } mathrm{kg})}=frac{0.2}{60} times frac{1000}{20} ) Putting the values in Eq. (i) ( therefore quad 0.45=left[1-frac{alpha}{2}right](5.12)left[frac{0.2}{60} times frac{1000}{20}right] ) ( 1-frac{alpha}{2}=frac{0.45 times 60 times 20}{5.12 times 0.2 times 1000} ) ( Rightarrow quad 1-frac{alpha}{2}=0.527 ) ( Rightarrow quad frac{alpha}{2}=1-0.527 ) ( therefore quad alpha=0.946 ) Thus, percentage of association ( =94.6 % )