1.2 mL of acetic acid is dissolved in water to make 2.0L 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 1.02g mL^-1

Molar mass of acetic acid is 60g mol^-1]

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Vishal Baghel | Contributor-Level 10

6 months ago

Mass = d * v = 1.02 * 1.2 = 1.224gm

Moles of acetic acid = 0.0204 moles in 2L

So molality = 0.0102 mol/kg

$Δ T_{f} = i * K_{f} * m$

i = 1 + a for acetic acid

0.0198 = (1 + a) * 1.85 * 0.0102

α = 0.04928 = 5%

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