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What is the boiling point of a solution made by mixing 75.0g ZnCl2 with 375.0 grams of water? (Kb for water is 0.512 C/m)

Respuesta :

tr847007
The problem can be solved using the following formula:

ΔTb = i Kb m

i = moles particles/moles solute
Kb = 0.512 °C/m
m = molality =  moles solute/kg solvent

First we can solve for the molality of the solution:

75.0 g ZnCl₂ / 136.286 g/mol = 0.550 mol ZnCl₂

m = 0.550 mol/0.375 kg
m = 1.468 mol/kg

We can now solve for the change in temperature of the boiling point:

ΔTb = i Kb m
ΔTb = (3 mol particles/1 mol ZnCl₂) (0.512 °C/m) (1.468 m)
ΔTb = 2.25 °C

The boiling point of a solution is the initial boiling point plus the change in boiling point:

BP = 100 °C + 2.25 °C
BP = 102.25 °C

The solution will have a boiling point of 102.25 °C.

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