Question

3. Arrange the five solutions in the order of increasing boiling point elevation, ∆Tb, and state your reason: (a) 0.25 m sucrose (aq) (b) 0.15 m KNO3(aq) (d) 0.048 m C10H8 (naphthalene) in benzene (d) 0.15 m CH3COOH(aq) (e) 0.15 m H2SO4(aq)

4. At a barometric pressure of 744mmHg, the boiling point of water is 99.4 C. What approximate mass of NaCl would add to 3.50 kg of the boiling water to raise the boiling point to 100.0 C?

Answer 1

The change in boiling point of a solution is always given by:
\[\Delta T{_b} = iK{_b}m\]
where i is the van'Hoff factor, Kb is the molal boiling point constant, and m is the molality (NOT molarity) of the solution. Assuming all 5 of those solutions have approximately equal Kb (and you'd have to look that one up), the difference comes down to the overall concentration and the van'Hoff factor.

The van'Hoff factor is also called the ion factor, and is a measure of the # of ions that a solute will create when it enters a solution. Both the sucrose and naphthalene are covalent, so their van'Hoff factors are both 1. The KNO3 and acetic acid form 2 ions when they dissociate, so their van'Hoff factors will both be about 2. The sulfuric acid forms 3 ions, so it will have the largest van'Hoff factor, probably around 2.6.

If all 5 solutions have roughly the same Kb values, the smallest change in BP will be the naphthalene, because it has the lowest molality (0.048m). Next will be the sucrose (0.25m). The acetic acid will probably be third, since both it and the KNO3 are ionic, but the acetic acid barely dissociates, so its van'Hoff factor will be lower than KNO3s. Next is the potassium nitrate, and the sulfuric acid should have the largest change in the BP.

Of course, if the Kb values are drastically different for each solute, then all this goes out the window and you'll just have to calculate each formula manually.