Calculate the freezing point (0°C) of a 0.05500 m aqueous solution of glucose. The molal freezing- point- depression constant of water is 1.86 °C/m.
A) - 0.05627
B) - 0.2046
C) 0.1023
D) 0.0286
E) - 0.1023

Question

Quizplus · Accepted Answer

First, we need to use the formula 
ΔTf = Kf x m
where ΔTf is the change in freezing point, Kf is the molal freezing point depression constant, and m is the molality of the solution.

We are given Kf = 1.86 °C/m and m = 0.05500 m. Plugging these values into the formula, we get:

ΔTf = (1.86 °C/m) x (0.05500 m) 
ΔTf = 0.1023 °C

Since the freezing point of pure water is 0°C, we can find the freezing point of the solution by subtracting the change in freezing point from the freezing point of pure water:

Freezing point = 0°C - 0.1023°C 
Freezing point = -0.1023°C

Therefore, the best choice is E: -0.1023.

Calculate the Freezing Point (0°C) of a 0