Question

The depression in freezing point of 0.2 m aqueous solution of an electrolyte is 0.7 K. Calculate its percent degree of dissociation. [K_f for water = 1.86 K kg mol^-1; n = 2]

पर्याय

• 29 %

• 46 %

• 60 %

• 88 %

Answer 1

88 %

Explanation:

Given: m = 0.2 m, K_f = 1.86 K kg mol^-1

Number of ions produced(n) = 2,

ΔT = 0.7K

ΔT_f = i × K_f × m

where i = 1+ α (α is degree of dissociation)

ΔT_f = (1 + α) × K_f × m

∴ 0.7 = (1 + α) × 1.86 × 0.2

0.7 = (1 + α) × 0.372

\[1+\alpha=\frac{0.7}{0.372}=1.882\]

∴ α = 1.882 – 1 = 0.882

Percent degree of dissociation = α × 100

= 0.882 × 100 

≈ 88%