Question

For most solvents the boiling point constant, Kbp, is less than the freezing point constant, Kfp. Give a thermodynamic explanation for this. (Hint: How can you calculate these constants from thermodynamic values?)

Answer 1

SOLUTION:

By additing a solute the boiling temperature of solvent increases, which is measured as a colligative property called elevation in boiling point.

Thus

TBP(soln) = TBP(pure) + delta T

and the boiling point is HIGHER than water.

The freezing temperature of a solution is less than that of pure solvent , and sis measured as depression in freezing point, a colligative property.

Thus TFP (sol) = TFP(pure) - delta T

and freezing point is lowered.

This is due to the colligative property of solvents .

The colligative property of elevation is boiling point is calculated using the expression

delta T (B) = i x m x K(fP)

K(fp)=deltaT(B)/ (ixm)

Similarly, deltaT(F)=i x m K(bp)

K(bp)=deltaT(F)/ (i xm)

where i = van't Hoff factor

m = molal concentration of solute

Kfp = freezing point constant for the solvent

Kbp= boiling point constant for the solvent