Calculate the boiling-point elevation for 2.0 kg of water containing 300. g of the salt CaCl2 (Kb = 0.512 C/m)

Answer 1

dT = i*Kb*m
i = 3 for CaCl2 since it gives three particles upon ionization
Kb = 0.512
m = mol/kg solvent
mols CaCl2 = 300 g/molar mass CaCl2 = ?
m = ?mols CaCl2/2
Substitute m, i, Kb and solve for the elevation.
Post your work if you get stuck.

Answer 2

dT = i*Kb*m
i = 3 for CaCl2 since it gives three particles upon ionization
Kb = 0.512
m = mol/kg solvent 
mols CaCl2 = 300 g/molar mass CaCl2 = ?
m = ?mols CaCl2/2
ANSWER: 1.32 degrees