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

💡 Solution & Explanation

0.85% means 8.5 g in 1 L. Molar mass of $\text{NaNO}_3 = 23 + 14 + 48 = 85 \text{ g/mol}$. Molarity $C = 8.5 / 85 = 0.1 \text{ M}$. $n=2$, so $i = 1 + \alpha = 1 + 0.90 = 1.90$. Osmotic pressure $\pi = iCRT = 1.90 \times 0.1 \times 0.0821 \times 300 = 1.90 \times 2.463 = 4.6797 \text{ atm} \approx 4.68 \text{ atm}$. Therefore, correct answer is B.