The boiling point elevation of an aqueous sucrose is found to be 0.39 Celsius. What mass of sucrose (molar mass = 342.30g/mol) would be needed to dissolve in 500.0 g of water? Kb (water) = 0.513 C/m

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

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