Since pure water boils at 100.00 C, and since the addition of solute increases b
ID: 1053200 • Letter: S
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+Tb)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=(0.00Tf)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.?
Explanation / Answer
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