A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 grams of water. What is the freezing point depression of the solvent if the freezing po

Question

A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 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.

Answer 1

The freezing point depression 

ΔT_f = k_fmi

where k_f = -1.86 for water

m = molality

i = number of particles (or ions) in solution

G = grams of solute

W = kilograms of solvent = 0.355

MM = molar mass of C₆H₁₂O₆ (Use a Periodic Table to determine this).

 

First find the moles of solute = G/MM

You can find the molality by

m = # moles of solute/ kilograms of solvent = (G/MM)/W

Since the solute is a molecular compound, it does not form ions in aqueous solution. Hence i = 1.

Substitute the values of k_f, m, i into ΔT_f = k_fmi