The vapour pressure of water is 12.3 kPa at 300 K. What is the vapour pressure of 1 molal aqueous so — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

A 1 molal solution contains 1 mole of solute in 1000 g water. Moles of water $N = 1000/18 = 55.55$. Mole fraction of solute $X_{solute} = 1 / (1 + 55.55) = 1/56.55 = 0.01768$. Relative lowering of vapour pressure is equal to mole fraction of solute. $P = P^\circ(1 - X_{solute}) = 12.3(1 - 0.01768) = 12.3(0.98232) = 12.082 \text{ kPa} \approx 12.08 \text{ kPa}$. Therefore, correct answer is B.