Lowering of vapour pressure in 1 molal aqueous solution at 100°C is — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

A 1 molal solution contains 1 mole of solute in 1000 g of water. Moles of water $= 1000 / 18 = 55.55$ mol. Mole fraction of solute $X_{solute} = 1 / (1 + 55.55) = 1/56.55$. At 100°C, the vapour pressure of pure water ($P^\circ$) is 760 mm Hg. The lowering of vapour pressure is $\Delta P = P^\circ X_{solute} = 760 \times (1/56.55) \approx 13.44$ mm Hg. Therefore, correct answer is A.