Chemistry question I need help with, please show your work!

Question

Determine the freezing point of a solution containing 50.0 g of sucrose dissolved in 750. g water; round to the nearest 0.01°C. (Kf for water is 1.86°C/m, and the molar mass of sucrose is 342.3 g/mol.)

Answer 1

Colligative Properties:

Freezing point depression...

∆T = imK

∆T = change in temperature = ?

i = van't Hoff factor = 1 for sucrose because it does not ionize or dissociate

m = molality = moles solute / kg solvent = 50.0 g sucrose x 1 mol/342.3 g / 0.750 kg = 0.1948 m

K = freezing constant = 1.86º/m

∆T = (1)(0.1948)(1.86)

∆T = 0.362ºC

Freezing point of solution will be -0.362ºC (since normal freezing point is 0ºC)