Question

0.5g KCl was dissolved in 100g water and the solution originally at 200C, froze at – 0.240C. Calculate the percentage ionization of salt.

Answer 1

Given - weight of KCl (w2) = 0.5 g

weight of water (w1)= 100 g

Tf = -0.240°C ,

kf for water = 1.86 k kg mol-1 .

Calculate- Percentage ionization of Salt.

Solution - We know that ,

Observed molecular mass , m

= Kf * w2 * 1000/ ∆Tf * w1

• ∆Tf = T°f - Tf = 0 - (-0.240) = 0.240 ° C

• m = 1.86 * 0.5 * 1000

0.240 * 100

= 38.75 g .

• But normal molecular mass of KCl = 74.5 g.

• So , Van't Hoff factor i = normal molecular mass/ observed molecular mass.

= 74.5/38.75

= 1.92

• Now , for finding percentage of ionization of Salt , we should know about the ionization of KCl , so it dissociates as _

KCl —> K+ + Cl-

moles after 1-α α α

dissociation

• Total number of moles after dissociation

= 1-α+α+α

= 1+α

• As, we know i = observed moles of solute

normal moles of solute

i = 1+α / 1

1.92 = 1+ α

α = 1.92-1 = 0.92

• Degree of dissociation or percentage of ionization of Salt = 0.92 *100 = 92% .