Question

What is the boiling point of a solution produced by adding 390 g of cane sugar (molar mass 342.3 g/mol) to 2.0 kg of water? For each mole of nonvolatile solute, the boiling point of 1 kg of water is raised 0.51 ∘C.

Answer 1

Lets calculate molality first

mass of solute = 390 g

we have below equation to be used:

number of mol of solute,

n = mass of solute/molar mass of solute

=(390.0 g)/(342.3 g/mol)

= 1.139 mol

mass of solvent = 2.0 Kg

we have below equation to be used:

Molality,

m = number of mol / mass of solvent in Kg

=(1.139 mol)/(2.0 Kg)

= 0.5697 molal

lets now calculate deltaTb

deltaTb = Kb*m

= 0.51*0.5697

= 0.2905 oC

This is increase in boiling point

boiling point of pure liquid = 100.0 oC

So, new boiling point = 100 + 0.2905

= 100.29 oC

Answer: 100.29 oC