UPSEE Solved Model Paper 5

Given
Molarity m = 0.5
Degree of ionization α = 20% = 20/100 = 0.2
Freezing point constant Kr = 1.86 K kg mol−1
Lowering in freezing point ΔTf = ?
cant Hoff's factor , i = Total no of mole sat equilibrium / initial moles
The given reaction is
HX → H++X−

I	0.5	0	0
C	c−cα	cα	cα

α = 0.2 , c = 0.5 m
c - cα = 0.4 M , cα = 0.1 M
so total moles are equilibrium = c - cα + cα + cα = 0.4 + 0.1 + 0.1 = 0.6 M
so, vant hoff factor i = 0.6/0.5 = 1.2
We have the equation
Lowering in freezing point ΔTf = iKfm
= 1.2 × 1.86 × 0.5 = 1.12 K