Question

assuming 100% dissociation, calculate the freezing point and boiling point of a 2.5 m SnCl4 (aq)

Answer 1

100% dissociation means you have Sn^4+ + 4Cl^-, giving you an i (or van't Hoff factor) of 5. You're in an aqueous solution, so we know that water freezes at 0C, and boils at 100C. The equation you're looking for is:
\[\Delta T = iK _{f}m\]
Where i = van't hoff factor, Kf = freezing point depression constant, and m = molality.
Kf is a constant, 1.86 °C kg/mol
i is 5 as stated above
molality....
2.5 moles SnCL4 per 1 L water, want molality (mol solute/kg solvent)

1L water = 1000g at standard temp/pressure. 1.000kg

2.5 mol SnCl4/1.000kg H2O = 2.5m

Temp change = 5 (unitless) * 25 mol/kg * 1.86 °C kg/mol
Temp change = 23°C
So the freezing point is now -23°C because adding solutes lowers the freezing point.

Boiling point, same business except the constant changes, as does whether you add/substract your change in temperature.

\[\Delta T = iK _{b}m\]
Kb = 0.512 °C kg/mol

Temp change = 5 * 2.5 mol/kg * 0.512 °C kg/mol = 6.4 °C
Your new boiling point is 106°C. Viola! Hope this helped.