KCET 2014 Chemistry Solved Paper

Given,
wB=1.25g
wA=mass of solvent = 50 g
Tf=0.3°C
Molecular mass of P = 94
Kf=1.86K kg mol−1
ΔTf=0−(−0.3)=0.3°C
MB=‌

Kf×wB×1000

∆Tf×wA

MB=‌

1.86×1.25×1000

0.3×50

=155
Now, van’t Hoff factor,
i=‌

‌ Normal molar mass ‌

‌ Observed molar mass ‌

=‌

94

155

=0.6064
           P ⇌ P2
Initial moles     ‌‌‌‌1  0
Moles after association 1 − α α/2
(If α is the degree of association)
Degree of association (α) =‌

n(1−i)

n−1

=‌

2×(1−0.6064)

1

=78.7%