A beaker (A) contains 20 g sugar in 100 g water. Another beaker (B) contains 10 g of sugar in 100 g — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

At equilibrium, both solutions must have the same vapour pressure, meaning their concentrations (mass ratio of solute to solvent) must be identical. Total sugar = 30 g. Total water = 200 g. Let mass of water in A be $W_A$. Then $20 / W_A = 10 / (200 - W_A) \Rightarrow 2(200 - W_A) = W_A \Rightarrow 400 - 2W_A = W_A \Rightarrow 3W_A = 400 \Rightarrow W_A = 133.33 \text{ g}$. Initial water in A was 100 g, so $133.33 - 100 = 33.33 \text{ g}$ of water transfers from B to A. Therefore, correct answer is C.