Question

5% aqueous solution of kcl was found to freeze at – 0.24°c. Calculate the van't hoff factor and degree of dissociation of the solute at this concentration (kf for h2o = 1.86 k kg mol–1).

Answer 1

Answer:

ΔTf=i×kf×w2×1000Mw2×W1

0.24=i×1.86×0.5×100074.5×99.5⇒i=1.92

Total number of particles =1−α+α+α=1+α

i=1+a

1.92=1+α

So, α=1.92−1=0.92

i.e.,92%dissociated.

Explanation:

Answer 2

Answer:

ΔT

t

=iK

f

×m

(273−272.76)=i×1.86×

99.5

0.5/74.5

×1000

i=

500×1.86

0.24×99.5×74.5

=1.92

∴ Vant-Hoff factor =i=1.92

KCl⇋ K

+

+Cl

−

1 0 0 : Initial

1−x x x: At eqbm

i=

1

1+x

=1+x=0.92 ⇒x=0.92