6 . A
This is an example of reverse-phase chromatography. The solvent system is polar, which means
that the most polar compound will travel the furthest up the card, resulting in the largest Rf. This
gives compound III the largest Rf, which corresponds to spot A.7 . B
Because methylene chloride is denser than brine (salt water), the organic layer will settle at the
bottom of the funnel. Methylene chloride is nonpolar, so it cannot mix with brine, eliminating
choice (C).8 . B
Silica gels are polar. Polarity is used to selectively attract specific solutes within a nonpolar
solvent phase. Although silica gels have other properties, this is the most important to TLC.9 . B
In this question, three substances must be separated using a combination of techniques. The first
step should be the most obvious: remove the sand by filtration. The remaining compounds—
benzoic acid and naphthalene—are still dissolved in ether. If the solution is extracted with
aqueous base, the benzoate anion is formed and becomes dissolved in the aqueous layer, while
naphthalene, a nonpolar compound, remains in the ether. Finally, evaporation of the ether will
yield purified naphthalene.10 . C
This is the only option that would be effectively separated by a simple distillation. Choice (B)
would require vacuum distillation because the boiling points are over 150°C. In choices (A) and
(D), the boiling points are within 25°C of each other and would therefore require fractional
distillation in order to be separated.
11 . C
Affinity chromatography, using the target for the biological effector or a specific antibody, would
work best in this case. It will specifically bind the protein of interest and keep it in the column.
12 . D
Because this solution is composed of a much larger molecule and a much smaller molecule, size-
exclusion chromatography would effectively remove the smaller insulin molecule into the
fraction retained in the column and allow the titin to be eluted. Affinity chromatography, choice