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If 6 moles of sugar was added to a kilogram of water, the new boiling point would be _____ degrees C.

User Ben Carey
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Using the equation for boiling point elevation Δt
Δt = i Kb m
we can find the new boiling point T for the solution:
Δt = T - 100∘C
since we know that pure water boils at 100 °C.

We know that the van't Hoff Factor i is equal to 1 because sugar does not dissociate in water.

Also, the value of Ebullioscopic constant Kb for water is listed as 0.512 °C·kg/mol.

The molality m of the solution of 6 moles of sugar dissolved in a kilogram of water can be calculated as
m = 6 moles / 1 kg
= 6 mol/kg

Therefore the new boiling point T would be
T - 100 °C = i Kb m
T = i Kb m + 100 °C.
= (1) (0.512 °C·kg/mol) (6 mol/kg) + 100 °C
= 3.072 °C + 100 °C
= 103.072 °C

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