UPSEE Solved Model Paper 4

Given to (boiling point of benzene) = 80.10 °C = 353.1K molality (m) = 0.5
Δvap H = 30.775 KJ/mol
MW1 (benzene) = 78
Kb =

RT02MW1

1000ΔvapH

=

8.314×(353.1)2×78

1000×30.775×103

= 2.63
ΔTb = IKbm
⇒ I =

ΔTb

Kbm

=

0.7

2.65×0.5

i = 0.532
Considering association of acetic acid
2CH3COOH ⇌ (CH3COOH)3
Initialy : 1 , 0
At equilibrium : 1 - α , λ/2
Total number of moles = 1 - α +

α

2

= 1 -

α

2

i =

Total‌number‌of‌moles

Initial‌moles

0.532 = 1−

α

2

α = 0.936
% association = 93.6 %
Hence, option C is correct.