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Verified Answer :

meerkat18

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: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:

\(\Delta T_f=i\times K_f\times m\)

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

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

\(K_f\)= freezing point constant= 1.86Kkg/mol

\(273 K-T_f=5\times 1.86 Kkg/mol\times 2.5\)

\(273 K-T_f=23.25\)

\(T_f= 249.75K\)

The freezing point of the solution is \(249.75K\).

b) Elevation in boiling point:

\(\Delta T_b=i\times K_b\times m\)

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

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

\(K_b\)=boiling point constant= 0.512 Kkg/mol

\(T_b-373K=5\times 0.512 Kkg/mol\times 2.5\)

\(T_b-373K=6.4\)

\(T_b=379.4K\)

The boiling point of the solution is \(379.4K\).

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