Question

Calculate the freezing point and boiling point of each of the
following solutions: the boiling point of the solution: 60.0 g of
glucose, C_{6}H_{12}O_{6}, added to 113 g
of water (K_{b}=0.52∘C) Express your answer using one
decimal place.

Answer #1

Boiling point depression delta Tb is given as :

Delta Tb= kb.m

where kb is the boiling point constant and m is the molality

m = number of moles / mass of solvent in kg

m = (60/180.1559) / 0.113

m = 2.95m

Delta Tb = 0.512 x 2.95

Delta Tb = 1.51

So the boiling point of solution will be 100 + 1.51

= 101.51^{o}C

Freezing point depression of the solution Delta Tf = kf.m

putting the values we get :

Delta Tf = 1.86 x 2.95

= 5.5^{o}C

So the freezing point of the solution will be 0^{o} -
5.5^{o}

= -5.5^{o}C

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