Question

A 0.5 percent aqueous solution of Potassium Chloride was found to freeze at -0.24°C.
Calculate the Van't Hoff factor and degree of dissociation of the solute at this concentration
(K for water is 1.86).​

Answer 1

∆T = i×1000×Kw/W.M

0.24 = (i×1000×0.5×1.86)/100×74.5

i = (0.24×100×74.5)/(1000×0.5×1.86)

i = 1.923

i = 1 + (n-1)a

1.923 = 1 + (2-1)a

1.923 = 1 + a

a = 1.923 - 1

a = 0.923

Hence, Van't Hoff factor (i) = 1.923

and, Degree of Dissociation (a) = 0.923

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

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