Question

The boiling point elevation of an aqueous sucrose solution is found to be 0.39°C.

Answer 1

Answer:

130.4 grams of sucrose, would be needed to dissolve in 500 g of water.

Explanation:

Colligative property of boiling point elevation:

ΔT = Kb . m . i

In this case, i = 1 (sucrose is non electrolytic)

ΔT = Kb . m

0.39°C = 0.512°C/m . m

0.39°C /0.512 m/°C = m

0.762 m (molality means that this moles, are in 1kg of solvent)

If in 1kg of solvent, we have 0.712 moles of sucrose, in 500 g, which is the half, we should have, the hallf of moles, 0.381 moles

Molar mass sucrose = 342.30 g/m

Molar mass . moles = mass

342.30 g/m . 0.381 m = 130.4 g