Eliminate (A) and (C) since each addresses the canopy layer and the question is about the
other two. Choice (B) is disproven by Figure 3, and (D) is proven by Figures 1 and 3.
Passage IV
18 . F Use Figure 1. The key in the corner of the graph says that the anything graphed with a solid
line is blue, and anything graphed with a dotted line is colorless. The curve shown in this
graph changes from solid to dotted at 1 mL of titrant added, meaning that all the solution at all
values less than 1 mL will be blue, and the solution at all values greater than 1 mL will be
colorless. The only one of the answer choices that gives a value less than 1 mL is (F).
19 . D The blurb contains the following sentence: Starch and I 2 form a complex with a deep blue
color, but when I 2 is reduced to 2 iodide (I−) ions, the complex dissipates and the solution
becomes colorless. In other words, when the solution is reduced, it becomes colorless.
According to Figure 2, the solution is colorless (and therefore reduced) above 1 mL of titrant
added. Only (D) gives a value greater than 1 mL.
20 . J Figure 2 does not show the voltage at 2.5 mL of titrant added, but the curve follows a clear
trend. As the volume of titrant added increases, the voltage increases as well. At 2 mL of
titrant added, the voltage is equivalent to 7 kV. Therefore, at 2.5 mL of titrant added, the
voltage will most likely be greater than 7 kV, as in (J).
21 . C The experiment detailed in this passage is described in the first line as follows: Oxidation-
reduction titration is a method in which precise volumes of a titrant (an oxidizing or
reducing agent) are added dropwise to a known volume of an analyte (a reducing or
oxidizing agent, respectively). In other words, one substance (the titrant) is added gradually
to a certain amount of another substance (the analyte). Therefore, when Experiment 1 says
that the I 2 solution was incrementally added to the SO 2 solution, it can be inferred that the I 2
is the titrant, and the SO 2 is the analyte. In Experiment 2, the sodium thiosulfate solution was
used instead of the SO 2 solution; therefore, in Experiment 2, the I 2 is still the titrant, and the
sodium thiosulfate solution is the analyte, as in (C). If you picked (D), be careful—you may
not have noticed the change from Experiment 1 to Experiment 2.
22 . G Use POE. Experiment 1 contains the following information: A potentiometer, which acts as a
control input for electronic circuits, was placed in the solution. The key word here is
electric currents. There’s nothing to suggest that the potentiometer has anything to do with
concentration, eliminating (F), or freezing or boiling point, eliminating (H) and (J). Only
(G) contains any reference to electric currents and is therefore the best answer.
