An aqueous solution (0.85%) of NaNO3 is apparently 90% dissociated at 27°C. The osmotic pressure of — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

0.85% (w/v) NaNO3 = 0.85 g in 100 mL => C = (0.85/85) * (1000/100) = 0.10 M. NaNO3 → 2 ions (n=2), α = 0.90 => i = 1 + 0.90 = 1.90. π = i * C * R * T = 1.90 * 0.10 * 0.0821 * 300 = 4.68 atm.