An Experienced Approach to an SAT Chemistry Problem

Like many SAT subject tests, the subject test in Chemistry is as much a fight against time and careless mistakes as it is an opportunity to demonstrate mastery in Chemistry. The test covers concepts that would be addressed during a full year college-prep chemistry course in high school – though not at the same depth. The Test

There are 85 questions to be answered in an hour, meaning a student needs to average about 42 seconds per question. Because of these time constraints, as well as the inability to use a calculator, one’s approach to even simple problems is critically important. There are many types of SAT Chemistry problems – and, as one would expect, many ways to approach different problems. One advantage of using a tutor to study is that the tutor (if they are of high quality) has seen thousands of chemistry problems, and will often reveal a much faster way of completing a problem than someone who has simply taken a couple semesters in chemistry.

To illustrate an expert test prep tutor’s approach to SAT Chemistry problems, we will use the following example.

A SAT Chemistry Problem and One Approach to Solving

178 mL of water is added to 66 mL of a 7.5-M solution of H2SO4. The final concentration of the solution is:

a. 11M
b. 14M
c. 0.07M
d. 6.5M
e. 2.03M

Most inexperienced students will immediately begin crunching the numbers. In this case, it’s not a very difficult calculation. The steps are as follows:

1. Convert 66 mL into 0.066 L, by dividing by 1000.

2. Multiply 0.066L by 7.5 M (mols/liter) to solve for the number of Mols of H2SO4 in the solution: 0.495 mol. (But as there is no calculator allowed, they will need to do that by hand.)

3. Now they might add the initial volume (66mL) to the added volume of water (178mL) to get the total final volume 244mL.

4. They will convert 244mL into 0.244L by dividing by 1000.

5. The student will then divide the 0.495 mols/0.244L to get 2.0287M as the final molarity. (They could get away with estimating at this point.)

6. The student sees this answer is closest to answer choice E, picks it and moves on.

Without a calculator, however, even the estimations for the above values will require substantial concentration. Additionally, the math described above is not easy and the probability of making a careless mistake is high. The student may get to the right answer – but only after taking few seconds to decide whether to long divide or estimate.

Should the student decide to estimate, the student may experience anxiety as the given numbers are not the easiest to manipulate. The time spent on the calculations and the
estimation is a valuable resource, and so is the concentration required. With so many problems over the course of the test hour, mental fatigue can prove to be a major factor in sub-optimal performance.

An Advanced Approach to Solving SAT Chemistry Problems

A more experienced test-taker will likely do one estimation, combined with a process of elimination based on simple chemistry.

For instance, here are the steps I would counsel that student take:

1. Notice that the problem is adding water. Concentration, or molarity, is a measure of the amount of a particular substance in solution, and is calculated by dividing the moles of the solute by the volume of the solution (in Liters): thus, Moles/Liters. Since we are adding water (which the problem tells us is the solvent), the denominator is increasing, and thus the concentration in general is decreasing. This makes sense from experience, for instance when putting milk in coffee. We dilute the milk in coffee by adding more coffee. So the molarity is going down, and thus we can eliminate answer choices A and B immediately because they imply that the molarity is increasing.

2. Notice that the volume added is more than the initial value but not close to 9x or 10x the initial value. This means the concentration is going to go down by more than half, but not by as much as 10x. (This is the only estimation; what follows are easy inferences based on that estimate.)

3. Since we know that Molarity (concentration) behaves in a somewhat linear fashion (since Molarity is a simple fraction Mols/Liters);
a. Answer choice C would require increasing the volume by a factor of 10. As previously noted, the added volume isn’t even close, so choice C eliminated.
b. Answer choice D would require less than doubling the volume of the solution. As previously noted, the volume is more than doubling, and so the concentration will at least be cut in half. Answer choice D is eliminated.

4. Select answer choice E and move on.

In brief, the experienced approach to this problem uses knowledge of chemistry, one easy estimation, and some streamlined deductions to eliminate all but the right answer with speed and ease. This is just one of many strategies that an experienced SAT Chemistry tutor can teach to a student, allowing them to optimize their performance and translate their knowledge into a high score.

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