Asked by bobbyjoe
Calculate the boiling-point elevation for 2.0 kg of water containing 300. g of the salt CaCl2 (Kb = 0.512 C/m)
Answers
Answered by
DrBob222
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.
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.
Answered by
Bob
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
ANSWER: 1.32 degrees
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