Question

To use freezing-point depression or boiling-point elevation to determine the molal concentration of a solution.

The freezing point, Tf, of a solution is lower than the freezing point of the pure solvent. The difference in freezing point is called the freezing-point depression, ΔTf:

ΔTf=Tf(solvent)−Tf(solution)

The boiling point, Tb, of a solution is higher than the boiling point of the pure solvent. The difference in boiling point is called the boiling-point elevation, ΔTb:

ΔTb=Tb(solution)−Tb(solvent)

The molal concentration of the solution, m, is directly proportional to ΔTf and ΔTb:

m=moles of solutekilograms of solvent

B) A solution of water (Kf=1.86 defreeC/m) and glucose freezes at − 3.75 degree C. What is the molal concentration of glucose in this solution? Assume that the freezing point of pure water is 0.00 degree C

C)

A solution of water (Kb=0.512 ∘C/m) and glucose boils at 101.56 ∘C. What is the molal concentration of glucose in this solution? Assume that the boiling point of pure water is 100.00 ∘C.

Answer 1

B) A solution of water (Kf=1.86 defreeC/m) and glucose freezes at − 3.75 degree C.

What is the molal concentration of glucose in this solution? Assume that the freezing point of pure water is 0.00 degree C

dTf = -Kf*m

m = dTf/Kf = (-3.75)/(-1.86) = 2.0161 molal

C)

A solution of water (Kb=0.512 ∘C/m) and glucose boils at 101.56 ∘C. What is the molal concentration of glucose in this solution? Assume that the boiling point of pure water is 100.00 ∘C.

dTb = Kb*m

m = dTb/Kb = (101.56-100)/(0.512) = 3.046875 molal