Question

A solution is made by dissolving 25.5 grams of glucose (C6H12O6) in 398 grams of water. What is the freezing point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem. 

Please help!

Answer 1

The freezing point depression is a coligative property.

The formula for the freezing point depression is:

ΔTf = Kf * m

Where m is the molality and Kf is the cryoscopic constant.

Kf of water is 1.86 °C / m

m = moles of solute / kg of solvent

moles of solute = mass in grams / molar mass

molar mass of C6H12O6 = 6*12g/mol + 12*1g/mol + 6*16g/mol = 180 g/mol

moles of C6H12O6 = 25.5 g / 180 g/mol = 0.14167 mol

m = 0.14167 mol / 0.398 kg = 0.3559 m

=> ΔT = 1.86°C/m * 0.3559m = 0.66°C

Answer: 0.66°C