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
Question
A beaker (A) contains 20 g sugar in 100 g water. Another beaker (B) contains 10 g of sugar in 100 g water. Both the beakers are placed under a bell jar and allowed to stand until equilibrium is reached. How much water will be transferred from one beaker to the other? Neglect the mass of water vapour present at equilibrium.
💡 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.