Question

1.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid is 0.0198°C. The percentage of dissociation of the acid is ____ . (Nearest integer)
[Given: Density of acetic acid is 1.02 g mL–1..
Molar mass of acetic acid is 60 g mol–1
Kf(H2O) = 1.85 K kg mol–1].

Answer 1

The correct answer is: 5

ΔTb = i × Kb × m

$=\frac{1.2×1.02}{60}$

As moles of solute are very less.

So, take molarity and molality same.

$0.0198=i×1.85×1.2×\frac{1.02}{60×2}$

$i = 1.05$

$α=\frac{i-1}{n-1}=\frac{0.05}{1}=0.05$