# Chemistry hw

posted by
**Akle** on
.

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.