To use freezing-point depression or boiling-point elevation to determine the mol

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, Tf:

Tf=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, Tb:

Tb=Tb(solution)Tb(solvent)

The molal concentration of the solution, m, is directly proportional to Tf and Tb: m=moles of solutekilograms of solvent

Quantitatively, the freezing-point depression Tf of a solution is related to the molality m and the freezing-point-depression constant Kf of the solvent by the equation

Tf=Kfm

where the freezing-point depression is the difference between the freezing points of the pure solvent and the solution.

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

Similar to the freezing-point depression, the boiling-point elevation Tb of a solution is quantitatively related to the molality m and the boiling-point-elevation constant Kb of the solvent by the equation

Tb=Kbm

where the boiling-point elevation is the difference between the boiling points of the solution and the pure solvent.

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.

Explanation / Answer

we know that

depression in freezing point is given by

dTf = kf x m

now

dTf = freezing point of pure solvent - freezing point of solution

so

dTf = 0 - (-3.75)

dTf = 3.75

also

given

Kf = 1.86

so

dTf = kf x m

3.75 = 1.86 x m

m= 2.106

so

the molal concentration of glucose is 2.106

2)

now

elevation in boiling point is given by

dTb = kb x m

now

dTb = boiing poing of solution - boiling point of pure solvent

using given information

dTf = 101.56 - 100

dTf = 1.56

also

kb = 0.512

so

dTb = kb x m

1.56 = 0.512 x m

m = 3.047

so

the molal concentration of glucose is 3.047

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.