At 293 K, the vapour pressure of water is 2400 Pa and the vapour pressure of an aqueous urea solutio — Solutions and Colligative Properties Chemistry Question
Question
At 293 K, the vapour pressure of water is 2400 Pa and the vapour pressure of an aqueous urea solution is 2300 Pa. The osmotic pressure (in atm) of solution at 300 K, if the density of solution at this temperature is $1185 \text{ kg/m}^3$, is ($R = 0.08 \text{ L-atm/K-mol}$)
💡 Solution & Explanation
Relative lowering of vapour pressure is $(2400 - 2300) / 2400 = 100 / 2400 = 1/24$. Thus, $X_{\text{urea}} = 1/24$ and $X_{\text{water}} = 23/24$. This ratio means 1 mole of urea (60 g) is dissolved in 23 moles of water ($23 \times 18 = 414 \text{ g}$). Total mass of solution = $60 + 414 = 474 \text{ g} = 0.474 \text{ kg}$. Volume $V = \text{mass} / \text{density} = 0.474 \text{ kg} / 1185 \text{ kg/m}^3 = 0.0004 \text{ m}^3 = 0.4 \text{ L}$. Osmotic pressure $\pi = (n/V)RT = (1 \text{ mol} / 0.4 \text{ L}) \times 0.08 \times 300 = 2.5 \times 24 = 60 \text{ atm}$. Therefore, correct answer is 0060.