Exam tomorrow...

Yesterday in class we (more or less) reviewed for the exam. We added normality to our list of concentrations, normality is closely related to molarity.

Since people didn't have any additional questions before the exam, we also started talking about kinetics, we'll dig deeper into rates starting Friday.

A few questions have come in:

How do we do a problem like this? (from Spring 2005 exam)
10. What is the mol fraction of sugar (C₆H₁₂O₆) in a saturated aqueous sugar solution at 25ºC?
(Solubility of sugar in water at 25ºC = 211.4 g/100.mL)
a. 0.1744
b. 0.6789
c. 11.73
d. 2.114
e. 0.2112

This is largely just a unit conversion problem. We're looking for mol fraction of sugar in water, so we need to know (mols of sugar), (mols of water), and (total mols). Looking at that solubility, we can convert 211.4g of sugar to mols (211.4g / 180.155g/mol = 1.1734mols sugar), and we can convert 100.mL of water to mols (100.g / 18.015g/mol = 5.5509mols water). The mol fraction of sugar is (1.1734mols sugar / (1.1734+5.5509)total mols = 0.1745) so the answer should be "a". Just a quick note on test strategy here, you know by definition that the only possible values of mol fraction are between 0 and 1; therefore, you know that "c" and "d" must not be correct. Even if you had to guess, eliminating those two impossible answers increases your odds from 20% to 33%.


Hi, for the Spring 2008 exam you gave, I was hoping you could post how to do number six regarding vapor pressure on the blog.
6. If each of the following solids is added to 500.0mL of water, which will change the vapor pressure the most?
a. 1.2mols sugar
b. 0.4mols calcium phosphate
c. 0.6mols sodium chloride
d. 0.5mols calcium nitrate
e. 0.7mols ammonium phosphate

We worked through this one in class on Monday. Since all of these solutes are being dissolved in the same amount of water, the one that provides the most solute particles will affect the vapor pressure the most. Sugar is a molecular solute, so 1.2mols of sugar will result in 1.2 mols of solute particles in solution. Calcium phosphate will break into 5 pieces, so 0.4mols of Ca₃(PO₄)₂ will result in 5(0.4) = 2.0mols of particles in solution. Sodium chloride = 2 particles, 2(0.6) = 1.2mols particles. Calcium nitrate = 3 particles, 3(0.5) = 1.5mols particles. Ammonium phosphate = 4 particles, 4(0.7) = 2.8mols particles. Since ammonium phosphate results in the most solute particles, it is expected to affect the vapor pressure the most.

Other questions, let me know.