Question

A solution is prepared by dissolving 16.90 g of ordinary sugar (sucrose, C12H22O11, 342.3 g/mol) in 40.90 g of water. Calculate the boiling point of the solution. Sucrose is a nonvolatile nonelectrolyte.

Answer 1

Answer:

Explanation:

The boiling point will increase due to dissolution of sugar in water . Increase in boiling point ΔT

ΔT = Kb x m , where Kb is molal elevation constant water , m is molality of solution

Kb for water = .51°C /m

moles of sugar = 16.90 / 342.3

= .04937 moles

m = moles of sugar /  kg of water

= .04937 / .04090

= 1.207

ΔT = Kb x m

= .51 x 1.207

= .62°C .

So , boiling point of water = 100.62°C .