Question

20 g of fructose (look up molar mass) is dissolved in 500 g of water. Calculate the boiling point

of this mixture.

Answer 1

Lets calculate molality first

Molar mass of C6H12O6 = 6*MM(C) + 12*MM(H) + 6*MM(O)

= 6*12.01 + 12*1.008 + 6*16.0

= 180.156 g/mol

mass of C6H12O6 = 20 g

we have below equation to be used:

number of mol of C6H12O6,

n = mass of C6H12O6/molar mass of C6H12O6

=(20.0 g)/(180.156 g/mol)

= 0.111 mol

mass of solvent = 500 g

= 0.5 kg [using conversion 1 Kg = 1000 g]

we have below equation to be used:

Molality,

m = number of mol / mass of solvent in Kg

=(0.111 mol)/(0.5 Kg)

= 0.222 molal

lets now calculate deltaTb

deltaTb = Kb*m

= 0.512*0.222

= 0.1137 oC

This is increase in boiling point

boiling point of pure liquid = 100.0 oC

So, new boiling point = 100 + 0.1137

= 100.114 oC

Answer: 100.114 oC