Calculate the freezing point of a 2.0 m solution of sucrose in water? Sucrose is a nonelectrolyte.The normal freezing point of pure water is 0.00$\degree$C.K_f(water) = -1.86$\degree$C/m A)-1.86 $\degree$C B)-3.72$\degree$C C)-2.00$\degree$C D)3.72$\degree$C E)1.86$\degree$C

Question

Quizplus · Accepted Answer

The freezing point depression is calculated using the formula ΔTf = i * Kf * m. For sucrose, i = 1 (nonelectrolyte), Kf = -1.86°C/m, and m = 2.0 m. So, ΔTf = 1 * (-1.86) * 2 = -3.72°C. The new freezing point is 0°C - 3.72°C = -3.72°C.

Calculate the Freezing Point of a 2 °\degree° CKf (water) = -1

Calculate the Freezing Point of a 2 $\degree$ CK_f(water) = -1