Question

Since pure water boils at 100.00 ∘C, and since the addition of solute increases boiling point, the boiling point of an aqueous solution, Tb, will be Tb=(100.00+ΔTb)∘C Since pure water freezes at 0.00 ∘C, and since the addition of solute decreases freezing point, the freezing point of an aqueous solution, Tf, will be Tf=(0.00−ΔTf)∘C

What is the boiling point of a solution made using 735 g of sucrose, C12H22O11, in 0.225 kg of water, H2O?

What is the freezing point of a solution that contains 28.2 g of urea, CO(NH2)2, in 255 mL water, H2O? Assume a density of water of 1.00 g/mL.?

Answer 1

1)

DTb = i*Kb*m

(Tb - T0) = DT

i = vanthoffs factor = 1

Kb of water = 0.512 K/m

m = molality = (w/mwt)*(1000/W)

mwt = molarmass of solute = 342 g/mol

w = weight of solute = 735 g

W = mass of solvent = 225 g

(Tb-0) = 1*0.512*(735/342)*(1000/225)

Tb = boiling point of solution = 4.9 C

2)

DTF = i*Kf*m

(T0-Ts) = DT

i = vanthoffs factor = 1

Kf of water = 1.86 K/m

m = molality = (w/mwt)*(1000/W)

mwt = molarmass of solute = 60 g/mol

w = weight of solute = 28.2 g

W = mass of solvent = 255 g

(0-Ts) = 1*1.86*(28.2/60)*(1000/255)

Ts = freezingpoint of solution = -3.42 C