An aqueous solution freezes at -.186c (Kf=1.86 ; Kb=.512c).what is the elevation in boiling point.?

Question

Answer

Explanation

Cryoscopic constant (Kf) and ebullioscopic constant (Kb) are the colligative properties of solutions. They depend only on the nature of the solvent and not on the nature of solute. The cryoscopic constant (Kf) is the depression in freezing point and the ebullioscopic constant (Kb) is the elevation in boiling point.

Calculation of Elevation in Boiling Point

The formula for calculating the elevation in boiling point is:

ΔTb = Kbm

where ΔTb is the elevation in boiling point, Kb is the ebullioscopic constant and m is the molality of the solution.

Since the solution freezes at -.186c, we know that the depression in freezing point is ΔTf = 0 - (-0.186) = 0.186c.

The molality of the solution can be calculated using the formula:

ΔTf = Kfm

where ΔTf is the depression in freezing point, Kf is the cryoscopic constant and m is the molality of the solution.

Substituting the given values, we get:

0.186 = 1.86 x m

m = 0.1 mol/kg

Now, substituting the calculated values in the formula for elevation in boiling point, we get:

ΔTb = Kb x m

ΔTb = 0.512 x 0.1

ΔTb = 0.0512c

Conclusion

The elevation in boiling point of the solution is 0.0512c. This means that the boiling point of the solution will be higher than that of the pure solvent by 0.0512c.

Answer

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