Question

What is the freezing point of a solution that contains 36.0 g of glucose (C6H12O6) in 500.0 g of water? (Kf for water is 1.86 C/m. The molar mass of glucose if 180.0 g per mole.)

Answer 1

-3.33 C

Step-by-step explanation:

Answer 2

Freezing point depression is a colligative property, which means that in only depends on the solute concentration.

The equation that relates freezing point depression with the solute concentration is:

ΔT = Kf * m

Where, ΔT is the freezing point depression, Kf is the cryoscopic constant and m is the molality or molal concentration.

Kf = 1.86 °C/m

m = moles of solute / Kg of solvent

moles of solute = mass / [molar mass] = 36.0 g / 180.0 g/mol = 0.200 mol

Kg of solvent = 500.0 g / 1000 g/kg = 0.5000 Kg of water

=> m = 0.200 / 0.5000 = 0.400 m

=> ΔT = 1.86 °C/m * 0.400m = 0.744 °C

Givent that the freezing point of pure water is 100°C, the new frezzing point is 100°C - 0.744 °C = 99.256°C.

Answer: 99.256 °C