chem homework question

Question

At , -12.5 ∘C, a common temperature for household freezers, what is the maximum mass of sucrose (C12H22O11) you can add to 1.50 kg of pure water and still have the solution freeze? Assume that sucrose is a molecular solid and does not ionize when it dissolves in water. 

Answer 1

Colligative Properties: Freezing point depression.

∆T = imK

∆T = change in freezing point temperature = -12.5ºC (since water normally freezes at 0ºC)

i = van't Hoff factor = 1 since sucrose does not ionize

m = molality = mols sucrose/kg water = ?

K = freezing constant for water = -1.86ºC/m

We will solve for m and then using 1.5 kg of water, we will find the moles of sucrose and then convert to mass.

m = ∆T/(i)(K) = -12.5º / (1)(-1.86º/m)

m = 6.72 mol/kg

Since we have 1.5 kg, we can find the moles of sucrose...

6.72 mol/kg x 1.5 kg = 10.08 moles sucrose

molar mass sucrose = 342 g/mol so we can now find mass of sucrose...

10.08 moles sucrose x 342 g/mol = 3447 g sucrose

Correcting for significant figures, we have 3450 g (3.45 kg) sucrose