JEE Main 2024 Chemistry question
Consider the dissociation of the weak acid HX as given below: HX(aq) ⇌ H⁺(aq) + X⁻(aq), = 1.2 × 10⁻⁵ : dissociation constant]. The osmotic pressure of 0.03 M aqueous solution of HX at 300 K is × 10⁻² bar (nearest integer). [Given: R = 0.083 L bar mol⁻¹ K⁻¹]
Answer: 76
Solution
α = √/c) = √(4 × 10⁻⁴) = 0.02, so i = 1.02. π = iCRT = 1.02 × 0.03 × 0.083 × 300 = 0.762 bar = 76 × 10⁻² bar.