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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 H 2 O = 1 . 85 K kg mol - 1 ]

Correct answer

0

Step-by-step solution

M = d × V = 1 . 02 × 1 . 2 = 1 . 224 gm Moles of acetic acid = 0 . 0204 moles in 2   L So molality = 0 . 0102   mol / kg Now ΔT f = i × K f × m i = 1 + α for acetic acid 0 . 0198 = 1 + α × 1 . 85 × 0 . 0102 α = 0 . 04928 ≅ 5 %

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