What is the freezing point of a solution that has 5.00 moles of NaI in 1250 g of water? (kf = 1.86oC/m)

Answer 1

Colligative property:

Freezing point depression

∆T = imK

∆T = change in freezing point temperature

i = van't Hoff factor = 2 for NaI b/c you get 2 particles when it dissociates

m = molal = moles NaI / kg solute = 5 mol / 1.250 kg = 4 m

K = 1.86 º/m

∆T = (2)(4)(1.86) = 14.9ºC

So new freezing point would be -14.9ºC