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 
K_f(H₂O) = 1.85 K kg mol^–1].

Answer 1

Correct Answer: 5

Given: Density of acetic acid = 1.02 g mL^–1, Volume of acetic acid = 1.2 mL.
We first calculate the mass of acetic acid:Mass = Density × Volume = 1.02 g mL^–1 × 1.2 mL = 1.224 g
Next, calculate the moles of acetic acid:Moles = Mass / Molar Mass = 1.224 g / 60 g mol^–1 = 0.0204 mol
Using the formula for depression in freezing point: ΔT_f = i×K_f×m
Where:ΔT_f = 0.0198°C,K_f = 1.85 K kg mol^–1,i = van 't Hoff factor, and m = molality.
Molality:m = Moles of solute / kg of solvent = 0.0204 mol / 2 kg = 0.0102 mol/kg
Plug values into the freezing point depression equation:0.0198 = i × 1.85 × 0.0102,i = 0.0198 / (1.85 × 0.0102) ≈ 1.06
The van 't Hoff factor, i, is related to the degree of dissociation (α) for acetic acid (CH₃COOH) as follows:i = 1 + α(A - 1), where A = 2, since dissociation gives two ions (CH₃COO^– and H⁺).
1.06 = 1 + α(2 – 1),α = 1.06 – 1 = 0.06 or 6%
Thus, the percentage of dissociation of the acetic acid is approximately 6%, which falls within the given range.