Question

A solution containing 0.5 g of KCl dissolves in 100 g of water and freezes at −0·24°C. Calculate the degree of dissociation of the salt. (K_f for water = 1.86°C) Atomic weights, K = 39, Cl = 35.5.

Answer 1

Given: Mass of KCl = 0.5 g

Mass of water = 100 g = 0.1 kg

Freezing point depression (ΔT_f) = 0.24°C

K_f = 1.86°C kg/mol

Molar mass of KCl = 39 + 35.5 = 74.5 g/mol

Molality (m) = `0.5/(74.5 xx 0.1)`

= `0.5/7.45`

= 0.0671 mol/kg

We know that

ΔT_f = i K_f m

`i = (Delta T_f)/(K_f * m)`

= `0.24/(1.86 xx 0.0671)`

= `0.24/0.1248`

= 1.92

Degree of dissociation for KCl

\[\ce{KCl -> K+ + Cl-}\]

i = 1 + α

1.92 = 1 + α

α = 0.92