Answer:
130 g of sucrose
Explanation:
Boiling point elevation formula → ΔT = Kb . m
ΔT = Boiling T° solution - Boiling T° pure solvent → 0.39°C
0.39°C = 0.513°C/m . M
m = 0.760 mol/kg → molality = moles of solute / 1kg of solvent
Let's determine the moles of solute → molality . kg
0.760 mol/kg. 0.5 kg = 0.380 moles
If we convert the moles to mass, we'll get the answer
0.380 mol . 342.30 g/mol = 130g
