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.

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.

Your approach is correct. Let's calculate the freezing point depression and boiling point elevation using the given information.

For freezing point depression:
ΔTf = i × Kf × m
ΔTf = 5 × 0.51°C/m × 3.13 m
ΔTf = 7.882°C

The freezing point is given by subtracting the ΔTf from the normal freezing point of water (0°C):
Tf = 0°C - 7.882°C
Tf = -7.882°C

For boiling point elevation:
ΔTb = i × Kb × m
ΔTb = 5 × 1.82°C/m × 3.13 m
ΔTb = 28.465°C

The boiling point is given by adding the ΔTb to the normal boiling point of water (100°C):
Tb = 100°C + 28.465°C
Tb = 128.465°C

Therefore, the freezing point (Tf) is -7.882°C and the boiling point (Tb) is 128.465°C.

Your calculations and approach are correct! You correctly identified that SnCl4 dissociates into 1 Sn 4+ ion and 4 Cl- ions, giving a van't Hoff factor (i) of 5. Using the formulas deltaT = i x K x m, where Kf is the freezing point depression constant and Kb is the boiling point elevation constant for water, you can calculate the freezing point depression (deltaTf) and boiling point elevation (deltaTb).

For the freezing point:
deltaTf = i x Kf x m = 5 x 0.51C/m x 3.13m = 7.96C

For the boiling point:
deltaTb = i x Kb x m = 5 x 1.82C/m x 3.13m = 28.28C

To find the new freezing point, subtract deltaTf from 0°C:
Tf = 0°C - 7.96°C = -7.96°C

To find the new boiling point, add deltaTb to 100°C:
Tb = 100°C + 28.28°C = 128.28°C

So, the freezing point (Tf) is -7.96°C and the boiling point (Tb) is 128.28°C for a 3.13 m SnCl4(aq) solution, assuming 100% dissociation.

If you want me to recheck your answers, please provide the values of Kf and Kb.