Assuming 100% dissociation, calculate the freezing point and boiling point of a 2.5 m sncl4 (aq)

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

Explanation :

Molality of = 2.5

a) Depression in freezing point:

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

= freezing point constant= 1.86Kkg/mol

The freezing point of the solution is .

b) Elevation in boiling point:

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

=boiling point constant= 0.512 Kkg/mol

The boiling point of the solution is .

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