A solution of sucrose (molar mass = 342 g/mol) has been prepared by dissolving 6.85 g of sucrose in 100 g of water. What is the freezing point of solution? (K_f for water = 1.86 K kg/mol)

- 0.372^oC

Depression in freezing point, ΔT_f = K_f x m

Molality, m = \(\frac{6.85/342}{100}\) x 1000 = 0.2

ΔT_f = 1.86 x 0.2 = 0.372

Freezing point of solution = 0 - 0.372 = -0.372 ^oC