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:

ΔT=iKfm

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.

ΔT=iKbm 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.

Answer 2

Answer:The freezing point of the solution is 249.75 K and boiling point of the solution is 379.4 K.

Explanation :

Molality of
`SnCl_4`= 2.5

a) Depression in freezing point:

`\Delta T_f=i* K_f* m`

`\Delta T_f=T^(o)_f-T_f=273 K-T_f`

i = Van'T Hoff factor = 5 (for 100% dissociation)

`K_f`= freezing point constant= 1.86Kkg/mol

`273 K-T_f=5* 1.86 Kkg/mol* 2.5`

`273 K-T_f=23.25`

`T_f= 249.75K`

The freezing point of the solution is
`249.75K`.

b) Elevation in boiling point:

`\Delta T_b=i* K_b* m`

`\Delta T_b=T_b-T^(o)_b=373 K`

i = Van'T Hoff factor = 5 (for 100% dissociation)

`K_b`=boiling point constant= 0.512 Kkg/mol

`T_b-373K=5* 0.512 Kkg/mol* 2.5`

`T_b-373K=6.4`

`T_b=379.4K`

The boiling point of the solution is
`379.4K`.