Respuesta :

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 [tex]SnCl_4[/tex]= 2.5

a) Depression in freezing point:

[tex]\Delta T_f=i\times K_f\times m[/tex]

[tex]\Delta T_f=T^{o}_f-T_f=273 K-T_f[/tex]

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

[tex]K_f[/tex]= freezing point constant= 1.86Kkg/mol

[tex]273 K-T_f=5\times 1.86 Kkg/mol\times 2.5[/tex]

[tex]273 K-T_f=23.25[/tex]

[tex]T_f= 249.75K[/tex]

The freezing point of the solution is [tex]249.75K[/tex].

b) Elevation in boiling point:

[tex]\Delta T_b=i\times K_b\times m[/tex]

[tex]\Delta T_b=T_b-T^{o}_b=373 K[/tex]

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

[tex]K_b[/tex]=boiling point constant= 0.512 Kkg/mol

[tex]T_b-373K=5\times 0.512 Kkg/mol\times 2.5[/tex]

[tex]T_b-373K=6.4[/tex]

[tex]T_b=379.4K[/tex]

The boiling point of the solution is [tex]379.4K[/tex].

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