What is the freezing point?

Question

What is the freezing point (in degrees Celcius) of 4.47 kg of water if it contains 226.7 g of 

CaBr2CaBr2? The freezing point depression constant for water is 1.86 ^oC/m and the molar mass of 

CaBr2CaBr2 is 199.89 g/mol

Answer 1

∆T = imK

∆T = change in freezing point = ?

i = van't Hoff constant = 3 for CaBr2 because it dissociates into 3 ions: Ca2+ and 2Br-

m = molality = moles solute/kg solvent = 226.7g/199.89/4.47kg = 0.2537 m

K = freezing point constant = 1.86º/m

∆T = (3)(0.2537)(1.86)

∆T = 1.42ºC

So the new freezing point will be -1.42ºC