What mass in grams of CH₃OH (MM = 32.04 g/mol) would need to be added to 460.1 g of water in order to lower the freezing point of water by 12.0 °C? (Kf for water is 1.86 °C/m)

Question

What mass in grams of CH₃OH (MM = 32.04 g/mol) would need to be added to 460.1 g of water in order to lower the freezing point of water by 12.0 °C? (Kf for water is 1.86 °C/m)

Answer 1

For the freezing point depression, we us the following equation:

∆T = imK

∆T = change in freezing point = 12.0º

i = van't Hoff factor = # of particles = 1 for CH3OH since it does not ionize or dissociate

m = molality = moles CH3OH/kg water = ?

K = 1.86ºC/m

Solving for m we have...

m = ∆T/(i)(K) = (12) / (1)(1.86

m = 6.45 molal = 6.45 moles CH3OH/kg

Since we have 460.1 g of water, that is 0.4601 kg

6.45 moles / kg x 0.4601 kg = 2.968 moles CH3OH needed

2.968 moles x 32.04 g/mol = 95.1 g CH3OH needed