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 : (Kf)water = 1.86 K kg mol-1 density of acetic acid is 1.2 g mL-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 CH3COOH CH3COO + H
Answer
19
Solution
Mass of acid = 5 × 1.2 = 6 g ⇒ moles = 660 = 0.1 mol
Water = 1 L = 1 kg ⇒ m = 0.1 mol kg-1 (C ≈ 0.1 M)
= KaC = 6.25 10-50.1 = 0.025
i = 1 + = 1.025 ∵ 1 particle → 2 ions
Tf = iKfm = 1.025 1.86 0.1
= 0.19 K ⇒ freezing point = -19 10-2 C
Answer: 19