Assuming 100% dissociation, calculate the freezing point and boiling point of 3.13 m SnCl4(aq).
Tf=?
Tb=?
My work: SnCl4 with 100% dissociation gives a van't hoff factor of 5, 1Sn 4+ ion and 4 Cl- ions
deltaT = i x Kf x m for freezing point = 5 x 0.51C/m x 3.13m = deltaTf
deltaT = i x Kb x m for boiling point = 5 x 1.81C/m x 3.13m = deltaTb
Kf for water = 0.51C/m
Kb for water = 1.82C/m
for new freezing point, 0 - deltaT
for new boiling point, 100 + deltaT
Is this right? Can you tell me your answer so I can compare it with mine. I have one more attempt. Thanks.
2 answers
What is the answer for Tb and Tf??????? please help.
You have Kf and Kb reversed. Kf for H2O is 1.86 and Kb for water is 0.51.
1 answer