Considering acetic acid dissociates in water, its dissociation constant is 6.25 × 10 –5. If 5 mL of — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

Mass of CH3COOH = V × d = 5 ml × 1.2 g/ml = 6 gm CH COOH n 0.1mol = = 3 CH COOH CH COOH 0.1 m M 0.1M ≈ = = CH3COOH CH3COO – + H + C C – Cα Cα Cα a C K α = −α 1 – α ≈ 1 ⇒ Ka = Cα –5 Ka 6.25 10 C 0.1 × α = = = 25 × 10 –3 V.f. (i) = 1 + α(n – 1) = 1 + α(2 – 1) = 1 + α = 1 + 25 × 10 –3 = 1.025 ∆Tf = iKfm = (1.025)(1.86)(0.1) = 0.19 = 19 × 10 –2