Considering acetic acid dissociates in water, its dissociation constant is 6.25 × 10⁻⁵. If 5 mL of acetic acid is dissolved in 1 litre water, the solution will freeze at -x × 10⁻² ° C, provided pure water freezes at 0 ° C.
x= _________. (Nearest integer)
Given : (K_f)_(water)=1.86 K kg mol-1 ; density of acetic acid is 1.2 g mol⁻¹ . ; molar mass of water =18 g mol⁻¹ . ; molar mass of acetic acid= 60 g mol⁻¹ . ; density of water =1 g cm⁻³
Acetic acid dissociates as CH₃ COOH ⇌ CH₃ COO⊖+H⊕

Solution

Chemistry ma’am · 4 steps

Get α from the dissociation constant, then let i = 1 + α scale the freezing point drop.

1. Moles and molality. Mass of acetic acid is 5 mL × 1.2 g mL⁻¹ = 6 g, which is 6/60 = 0.1 mol. One litre of water is 1 kg, so the molality is m = 0.1 mol kg⁻¹. Take the concentration as about 0.1 M too.
2. Degree of dissociation. Acetic acid is a weak acid, so Kₐ ≈ Cα². That gives α = √((6.25 × 10⁻⁵)/0.1) = √(6.25 × 10⁻⁴) = 0.025.
3. Van't Hoff factor. One CH₃COOH gives two particles, CH₃COO⁻ and H⁺. So i = 1 + α = 1.025.

   Δ T_f = i K_f m = 1.025 × 1.86 × 0.1 = 0.1907 K

4. So. The solution freezes at -0.1907 °C = -19.07 × 10⁻² °C, so x is 19.

The common slip: 16. Taking 5 mL of acid as 5 g, without using the density of 1.2 g mL⁻¹, gives Δ T_f ≈ 0.159 K and so 16, not 19.

The idea to keep. For a weak electrolyte in a colligative question, find α from Kₐ = Cα², then use i = 1 + (n-1)α, where n is the number of ions it splits into.

Answer from the NTA final answer key. Solution by PrepWiser.