J.R. S. answered 02/18/21
Ph.D. University Professor with 10+ years Tutoring Experience
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