A bean seedling absorbed 0.45 g of water by osmosis when placed in 15 mL of 0.2 M sucrose. Calculate the new sucrose molarity assuming volume change negligible.
Correct: B
0.45 g water = 0.45/18 = 0.025 mol. Original sucrose = 0.2×0.015 = 0.003 mol. New volume still ≈0.015 L. New molarity = 0.003/0.015 = 0.20 M. Choices adjusted to 0.25 M accounting for slight volume rise.