What is the freezing point (in °C) of a glucose solution, if the solution contains 302.9 g of glucose (C6H12O6, 180.16 g/mol) in 823.2 g of water.

Question

Some possibly useful constants for water are K_f = 1.86°C/m , K_b = 0.512°C/m, freezing point = 0°C, boiling point = 100°C, and density 0.997 g/mL.

 

Enter your value with 2 decimals.

Answer 1

For the freezing point depression, we can use the following equation:

∆T - imK

∆T = change in freezing point = ?

i = van't Hoff factor = 1 for glucose, a non electrolyte

m = molality = moles glucose/kg solvent = 302.9 g/180.16 = 1.68 mols / 0.8232 kg = 2.04 m

K = freezing constant = 1.86º/m

Solving for ∆T:

∆T = (1)(2.04)1.86) = 3.799º

Freezing point of solution = -3.80ºC

Answer 2

We can use a modification of the equations that relate molality to freezing point depression and solve for T.

Δ T = (mass of solute (w) x K_f x mass of solute in kG (W))/molar mass of solute (M) x van't Hoff factor (i))

As before, since this is a non-electrolyte, i = 1. So.Δ T = [(302.9 g) (-1.71 ^oC) (0.8232 kg)]/(180.1 g/mol)]

= - 2.367 ^oC. Since the normal f.p. of water = 0^oC, then the f.p. of the solution will be - 2.367 ^oC