MCAT Sample Questions : Physical Sciences

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MCAT Sample Questions : Physical Sciences

Passage V

A chemist performed the following experiments to investigate the melting and freezing behaviors of acetamide.

Experiment 1: Melting

A large beaker of water was heated to a slow boil. A thermometer was placed in a test tube and 10 g of acetamide crystals was added. The test tube was then lowered into the boiling water (100o C). The temperature was immediately read, and was reread every 15 sec. The acetamide was stirred before each reading. When the temperature reached 80o C, the acetamide started melting. After a period of time, when all the acetamide had melted, the temperature began to increase again. Results are shown in Figure 1.

Experiment 2: Freezing

Trial 1

The test tube from Experiment 1 was removed from the hot water and left to cool in air at 20o C. The temperature readings and stirring were continued every 30 sec. The temperature dropped to 80o C, where it remained constant. The acetamide slowly began freezing and was completely solid after 23 min. After this, the temperature again decreased. The time for freezing was considered to be excessive, so another trial was completed.

Trial 2

The same test tube was placed in boiling water until the acetamide was completely melted. For this trial, however, the test tube was then placed in a beaker of water at 20o C. The results are shown in Figure 1.

Figure 1 Melting and freezing behavior of acetamide

Some sample questions on this passage are as follows:

In Experiment 2, which of the following is the most important difference in the procedures used for Trials 1 and 2?
1. The amounts of acetamide used in each test tube
2. The surroundings that were used to cool the acetamide
3. The temperatures at which the trials were started
4. The lengths of time allowed for the acetamide to melt
Answer: B

Explanation: The only experimental difference in Trial 1 vs. Trial 2 is that, in Trial 2, the test tube is placed in water (20^oC) to cool rather than in air (also 20^oC). In other words, only the surroundings were different. Thus, answer choice B is the best answer.
During Experiment 1, which of the following would most likely have occurred if the water had only reached 90o C before the test tube was placed into it?
1. More water would have been needed to melt the acetamide.
2. Less water would have been needed to melt the acetamide.
3. The acetamide would not have melted.
4. The acetamide would have taken longer to completely melt.
Answer: D

Explanation: The melting point of acetamide is 80^oC; therefore, acetamide will melt when it is in a test tube that is placed in a water bath at 90^oC. The temperature of the water in the bath, not the amount of water in the bath, determines whether or not the acetamide will melt. The period of time for acetamide to melt, starting at 90^oC, is more than the corresponding period, starting at 100^oC (i.e., the temperature of boiling water). Thus, answer choice D is the best answer.
How are the designs of the two experiments important for producing useful results?
1. Both processes, melting and freezing, take place under controlled conditions.
2. Both processes, melting and freezing, take place without being controlled or monitored.
3. The amounts of acetamide are shown to control the temperatures of melting and freezing.
4. The amounts of acetamide are shown to control the times needed for melting and freezing.
Answer: A

Explanation: Without controlling the temperature (i.e., raising the temperature of the water bath above 80oC), the experimenter could not have observed melting or freezing. Without monitoring the time, the experimenter could not have determined the period of time for the samples to melt or freeze. The temperature of melting (freezing) of a pure substance such as acetamide is independent of the amount melted, and Experiment 2 shows that the surroundings control the period of time for freezing to occur. Thus, answer choice A is the best answer.
During Trial 1 of Experiment 2, if the temperature readings were taken at 1-min intervals instead of 30-sec intervals, the acetamide would most likely have become completely frozen at:
1. 11 min, 30 sec.
2. 23 min.
3. 46 min.
4. a lower temperature.
Answer: B

Explanation: The time period of melting is independent of the time intervals used by the experimenter to record temperatures. The sample would freeze completely after 23 min regardless of the time interval used by the experimenter to record temperatures. Thus, answer choice B is the best answer.
In Experiment 2, why was it necessary to place the test tube in hot water for Trial 2, in view of the fact that this was NOT done in Trial 1?
1. The water was boiling for Trial 1, but it needed to be cold for Trial 2.
2. The acetamide was cooled by air in Trial 1, but by water in Trial 2.
3. The temperature lowered more quickly for Trial 2 than it did for Trial 1.
4. The acetamide was liquid before Trial 1, but it was solid before Trial 2.
Answer: D

Explanation: After Experiment 1, the sample was removed from a hot water bath as a liquid. Subsequently, the sample froze during Trial 1. Therefore, the sample had to be reheated in a water bath above its melting point to start Trial 2 as a liquid. Thus, answer D is the best answer
If the data for Trial 1 were plotted in Figure 1, compared to the data for Trial 2, they would:
1. slope less steeply downward, and not all of the data could be shown.
2. slope more steeply downward, and all of the data could be shown.
3. slope upward, and all of the data could be shown.
4. slope upward, and not all of the data could be shown.
Answer: A

Explanation: If the data for Trial 1 were plotted, the temperature would drop to 80oC and remain at this melting temperature for 23 min (or 23 min x 60 sec/min = 1380 sec). The line at 80oC would not slope downward at all in the figure, and it would extend well past 270 sec, the maximum time shown in the figure. Thus, answer choice A is the best answer.

MCAT Sample Questions : Physical Sciences

Passage VI

The timbre, or quality, of a musical tone depends on the number and relative strengths of the harmonics including the fundamental frequency of the note. Figure 1a illustrates the first three harmonics of a tone. The addition of the first two harmonics is pictured in Figure 1b, and the addition of the first 3 harmonics is shown in Figure 1c.

Figure 1 Elements of a complex tone

The graphs in Figure 2 illustrate the characteristics of two adjacent tones from a bassoon. Figure 2a shows the pressure variations and the amplitudes of the harmonics for one of the tones, and Figure 2b shows the same information for the other tone

Figure 2 Pressure variations and amplitudes of harmonics for adjacent bassoon tones.

Following are some sample questions on this passage:

Which of the waveforms shown in Figure 1 has the shortest period?
1. First harmonic
2. Second harmonic
3. Third harmonic
4. The waveform in Figure 1c
Answer: C

Explanation: The tone with the shortest period has the shortest wavelength. In Figure 1a, the period of the third harmonic (the curve with the smaller dashes) is seen to be shorter than the other two harmonics. Thus, answer choice C is the best answer.
At the second position where the three curves intersect in Figure 1a, the curves are all:
1. in phase.
2. out of phase.
3. at zero displacement.
4. at maximum displacement.
Answer: C

Explanation: The three curves in Figure 1a intersect at three points in time. The second intersection occurs in the middle of the time axis. At that point all three curves have zero displacement. Therefore, answer choice C is the best answer.
If the frequency of the first harmonic in Figure 2a is 100 Hz, what is the period of the second harmonic?
1. 0.005 sec
2. 0.01 sec
3. 50.0 sec
4. 200.0 sec
Answer: A

Explanation: The period T and frequency f of a tone are related by T = 1/f. If the first harmonic has a frequency of 100 Hz, then the second harmonic has a frequency of 200 Hz. The period corresponding to 200 Hz is 1/200 s^-1 = 0.005 s. Thus, answer choice A is the best answer.
Which of the following graphs best illustrates the relative amplitudes of the harmonics in Figure 1?
Answer: A

Explanation: The amplitudes of the three harmonics can be compared in Figure 1a. The first harmonic is seen to be largest while the other two have equal amplitudes. Answer choice A best represents these observations.
If a fourth harmonic exists for the tone graphed in Figure 1, then, compared to the third harmonic, the fourth harmonic will have:
1. lower amplitude.
2. higher amplitude.
3. lower frequency.
4. higher frequency.
Answer: D

Explanation: A fourth harmonic would have a shorter period than the other three. Since T = 1/f, the fourth harmonic would have a higher frequency than the third harmonic. Therefore, answer choice D is the best answer.
The period of the waveform shown in Figure 1c is the:
1. same as the period of the first harmonic.
2. same as the period of the second harmonic.
3. same as the period of the third harmonic.
4. sum of the periods of the first, second, and third harmonics.
Answer: A

Explanation: The waveform in Figure 1c begins to repeat at the zero displacement point near the end of the time axis. This is the same time period as the first harmonic as seen in Figure 1a. Thus, answer choice A is the best answer.

MCAT Sample Questions : Physical Sciences

Passage VII

A student was asked to determine the identity of an unknown acid that was liquid at room temperature (20^o C). The student was told that the acid was one of those listed in Table 1.

Acid	Structure	Molecular weight (g/mole)	Melting point (^oC)	pK_a
Propionic	CH₃CH₂COOH	74.08	–21.5	4.88
Crotonic	CH₃CH=CHCOOH	86.09	71.6	4.69
Butyric	CH₃CH₂CH₂COOH	88.10	–7.9	4.82
Oxalic	HOOCCOOH	90.04	101	3.14 4.77

Table 1 Characteristics of Several Acids

The student added 0.22 g of the acid to 30.0 mL of H₂O(l). The student then titrated the solution with 0.10 M NaOH(aq) while monitoring the pH with a pH meter. The results are summarized in Figure 1.

Figure 1 Titration of the acid with 0.1 M NaOH(aq)

Based on the titration curve, the student proposed that the unknown acid had 1 –COOH group and a molecular weight between 85 and 92.

Following are some sample questions on this passage:

A comparison of which two compounds from Table 1 best shows the effect of molecular weight alone on melting point?
1. Propionic acid and crotonic acid
2. Propionic acid and oxalic acid
3. Propionic acid and butyric acid
4. Butyric acid and crotonic acid
Answer: C

Explanation: According to the data in Table 1, both structure and molecular weight (i.e., molar mass) affect the melting point of a compound. In order to assess the effect of molecular weight or mass alone, any other effects such as obvious structural differences must be minimized. This is best done by comparing two compounds that are structurally similar. Because the structures of propionic acid and butyric acid (Answer C) differ by only a CH₂ group, they best show that melting point increases with molar mass. All of the other answer choices compare two compounds that differ significantly in structure. Therefore, the melting points of these compounds include both molar mass and structural effects. Thus, answer choice C is the best answer.
Before titrating with NaOH(aq), what was the approximate H₃O⁺(aq) concentration of the solution containing the unknown acid?
1. 0.001 M
2. 0.01 M
3. 0.03 M
4. 0.3 M
Answer: A

Explanation: Figure 1 shows the pH of the solution to be about 3 before any NaOH(aq) is added. pH = -log[H₃O⁺] 3 = -log[H₃O⁺] [H₃O⁺] = 10^-3 M = 0.001 M = Answer A
The student prepared a 0.1 M aqueous solution of crotonic acid and a 0.1 M aqueous solution of oxalic acid, then adjusted the pH of each to 4.7 by adding NaOH. Which solution has a lower freezing point?
1. The crotonic acid solution, because it contains a lower molar concentration of solute particles
2. The crotonic acid solution, because it contains a greater percent mass of solute
3. The oxalic acid solution, because it contains a greater molar concentration of solute particles
4. The oxalic acid solution, because it contains a smaller percent mass of solute
Answer: X

Explanation: The freezing point depression of an aqueous solution is a colligative property (i.e., it depends on the number of solute particles in a given volume of water.) Given two solutions, the one with the greater number of solute particles per liter of solution freezes at the lower temperature. Answer C is the only answer that relates a larger number of solute particles directly to a lower freezing point. Oxalic acid is diprotic and ionizes in accord with the pK_a values in Table 1 to a greater extent than does crotonic acid. Subsequently, oxalic acid requires more NaOH than does crotonic acid to reach a pH of 4.7, and oxalic acid produces a larger number of particles in solution. Thus, answer choice C is the best answer.
During the titration summarized in Figure 1, the concentration of R–COOH equalled the concentration of R–COO^- when the pH approximately equalled which of the following? (Note: R is a hydrocarbon.)
1. 4.8
2. 6.2
3. 7.0
4. 9.2
Answer: A

Explanation: In a titration of R–COOH, the concentrations of R–COOH and R–COO- are equal at the mid-point of the titration. This is often called the half-equivalence point. From the expression for the equilibrium constant of a weak acid HA, when [HA] = [A^-], then [H₃O⁺] = K_a and pH = pK_a. Table 1 shows the pKa value for a monoprotic acid to be 4.69–4.88. Answer choice A (4.8) lies in this range, the other choices do not. Alternatively, Figure 1 shows the pH at the half equivalence point of a weak acid to be about 4.8. Thus, answer choice A is the best answer.
The student rejected crotonic acid as a possible identity of the unknown acid because crotonic acid:
1. is a strong acid.
2. is insoluble in H₂O.
3. is solid at room temperature.
4. has a molecular weight of 86.09
Answer: C

Explanation: The first sentence of the passage states that the unknown “was a liquid at room temperature (20^oC).” Table 1 shows that the melting point of crotonic acid is 71.6^oC, which means it is a solid at room temperature (i.e., it melts 51.6^oC above room temperature). Thus, answer choice C is the best answer.

MCAT Sample Questions : Physical Sciences

Passage VIII

Several features of sulfuric acid are given below.

Preparation of Sulfuric Acid

Sulfuric acid is commonly prepared by the combustion of elemental sulfur to sulfur dioxide, followed by the catalytic oxidation of sulfur dioxide to sulfur trioxide. Sulfur trioxide is then absorbed into a 98% aqueous solution of H₂SO₄, and water is added to maintain a 98% concentration. SO₃ reacts with the water in the aqueous solution according to Reaction 1.

SO₃(g) + H₂O(l) = H₂SO₄(l)

Reaction 1

Properties

Concentrated sulfuric acid is 98% H₂SO₄ and 2% water by mass. It has a density of 1.84 g/mL and a boiling point of 338^oC.

Preparation of Other Acids

HCl(g) and HNO₃(l) may be prepared by the reaction between sulfuric acid and the sodium salt of the corresponding conjugate base (Cl^- or NO^3-, respectively).

formation of SO₂

Sulfuric acid forms SO₂ gas when it reacts with several compounds. For example, I₂ and SO₂ are formed when I^- reacts with concentrated H₂SO₄; Br₂ and SO₂ are formed when Br- reacts with concentrated H₂SO₄. Cu⁺ and SO₂ are formed in hot solutions of Cu(s) in H₂SO₄. This last reaction is unusual, because most metals react with solutions of H₂SO₄ to form hydrogen gas and a metal sulfate.

Following are some sample questions on this passage:

When sulfuric acid reacts with copper, how does the oxidation number of the sulfur change?
1. From +4 to +6
2. From +6 to +4
3. From +6 to +8
4. From +8 to +6
Answer: B

Explanation: The passage states that sulfuric acid reacts with Cu(s) to produce Cu⁺ and SO₂. Thus, sulfuric acid is converted into sulfur dioxide, or H₂SO₄= SO₂. The oxidation number of sulfur in H₂SO₄ can be found by assigning oxidation numbers of +1 for hydrogen and -2 for oxygen. For the formula H₂SO₄ to be neutral, the sum of the oxidation numbers must be zero. If x is the oxidation number of sulfur in H₂SO₄, then: 2(1) + 4(- 2) + x = 0, and x = +6. Likewise, for SO2: 2(- 2) + x = 0, and x = +4. The change in oxidation number is from +6 to +4, which is answer choice B.
The apparatus shown below can be used to prepare HNO₃(boiling point = 86^oC)

The yield of HNO₃ collected in the tube can be maximized by maintaining the temperatures of the flask and tube, respectively, at:
1. 0^oC and 100^oC.
2. 100^oC and 0^oC.
3. 350^oC and 150^oC.
4. 350^oC and 100^oC
Answer: B

Explanation: The boiling point of HNO₃ is given in the question as 86^oC. Because HNO3 must boil out of the flask and be trapped in the tube, the temperature of the flask must be above the boiling point of HNO₃ (i.e., < 86^oC) and the temperature of the tube must be less than the boiling point of HNO₃ (i.e, < 86^oC). Answer B meets these criteria; the other answers do not. Thus, answer choice B is the best answer.
Which of the following is the balanced equation describing the combustion of elemental sulfur?
1. 2 H₂S + 3 O2 = 2 SO₂ + 2 H₂O
2. H₂S + 2 O₂ = SO₃ + H₂O
3. 2 SO₃ = 2 S + 3 O₂
4. S + O₂ = SO₂
Answer: D

Explanation: The combustion of elemental sulfur involves a reaction between oxygen (O₂) and sulfur (S). [Note: Though sulfur exists as S8 molecules, its reactions are normally written in terms of its empirical formula S.] Only Answer D shows such a reaction. Thus, answer choice D is the best answer.
In the second step of preparing H₂SO₄from elemental sulfur (the catalytic oxidation of SO₂), which strategy is most likely to increase the yield of SO₃ formed?
1. Reducing the reaction temperature
2. Reducing the reaction pressure
3. Removing SO₃ from the reaction mixture
4. Removing O₂ from the reaction mixture
Answer: C

Explanation: The reaction involves the formation of gaseous SO₃ from gaseous O₂ and gaseous SO₂.

O2(g) + 2 SO₂(g) = 2 SO₃(g)

According to Le Châtelier's principle, any action that causes the reaction to shift toward the right will cause O₂ and SO₂ to react and increase the yield of SO₃. Of the four possible actions (answers A-D), the removal of SO₃ as it forms will shift the reaction toward the right and is the most likely action to increase the yield of SO₃. Thus, answer C is the best answer.
If H₂(g) is formed from the reaction of Fe(s) with dilute H₂SO₄(aq), which species acts as the reducing agent?
1. Fe
2. FeSO₄
3. SO₄^2-
4. H₃O⁺
Answer: A

Explanation: The problem hypothesizes the liberation of hydrogen in accord with the unbalanced equation below.

Fe(s) + H₂SO₄ = H₂(g)

In the conversion, hydrogen goes from an oxidation state of +1 in sulfuric acid to 0 (zero) in H₂. Thus, the hydrogen in H₂SO₄ is reduced (i.e., it undergoes an algebraic decrease in oxidation state). Fe(s) is the reducing agent (i.e., it causes the reduction). Thus, answer choice A is the best answer.
Which of the following species has the smallest concentration in 98% H₂SO₄?
1. SO₄^2-
2. H₂SO₄
3. H₃O⁺
4. HSO₄-
Answer: A

Explanation: The first ionization of sulfuric acid, H₂SO₄, is normally 100% in water. However, under conditions of low water content, all of the H₂SO₄ cannot ionize. Qualitatively, the only source of SO₄^2- is HSO₄-. The Ka for the second ionization step of a parent acid is a few orders of magnitude smaller than that of the first step; therefore, SO₄^2- must be the least abundant species, because it is only formed in the second ionization step. Thus, answer choice A is the best answer.

Quantitatively, the mass relationship in 100 g of 98% H₂SO₄ is 98 g H₂SO₄ and 2 g H2O. The molar mass of H₂SO₄ is 98 g/mol and of water is 18 g/mol. Thus, in 100 g of 98% H₂SO₄ there is one mole of H₂SO₄ and 2/18 or 1/9 mole of H₂O. In excess water, H₂SO₄ would ionize completely. However, in this case (i.e., very low water content), only about 1/9 of a mole can react stoichiometrically with water to form H₃O⁺ and HSO₄-. Of this 1/9 mol, only a small fraction of the HSO₄- further ionizes to H₃O⁺ and SO₄^2-, because HSO₄^2- is a weak acid. Therefore, SO₄^2- is the chemical with the lowest concentration. Thus, answer choice A is the best answer.

MCAT Sample Questions : Physical Sciences

The standard potential for the reaction
Zn²⁺ + 2 e^- = Zn

equals –0.763 V. When a strip of Zn is placed in a 0.1 M solution of HCl(aq), does the zinc strip begin to dissolve?
1. Yes; H₂(g) and ZnCl(aq) are produced.
2. Yes; H₂(g) and ZnCL₂(aq) are produced.
3. No; no reaction occurs because Zn is less reactive than is H₂.
4. No; no reaction occurs because Zn²⁺ is less reactive than is H⁺.
Answer: B

Explanation: As written, the equation shows the reduction of Zn²⁺ to Zn. The negative reduction potential for Zn²⁺ means that Zn has a positive oxidation potential (i.e., Zn is easily oxidized.) Zinc will displace hydrogen, which has a zero standard reduction or oxidation potential (i.e., hydrogen is the standard against which other substances are measured.). As given in the problem, ionic zinc is Zn²⁺. Therefore, zinc metal liberates hydrogen gas and produces ZnCL₂(aq). Answer choice B is the best answer
Which of the following must be known in order to determine the power output of an automobile?
1. Final velocity and height
2. Mass and amount of work performed
3. Force exerted and distance of motion
4. Work performed and elapsed time of work
Answer: D

Explanation: Power is defined as the rate of doing work. For the automobile, the power output is the amount of work done (overcoming friction) divided by the length of time in which the work was done. Therefore, answer choice D is the best answer
In the molecule shown below, which bond is the longest?
1. A
2. B
3. C
4. D
Answer: B

Explanation: The bonds labeled C and D in the figure are of equal length but shorter than bond B. This is because two resonance structures can be drawn: in one resonance structure, bond C is a double bond, and in the second resonance structure, bond D is double bond. Double bonds (bond order = 2) are shorter than single bonds (bond order = 1), so the bond order for bonds C and D is about 1.5 and for bond B about 1. Bond B is longer than bond A because of the small atomic radius of hydrogen compared to nitrogen. Thus, answer choice B is the best answer.

Alternatively, the figure shows that C = D, so answer choices C and D can be eliminated. Also, A is clearly shorter than B; therefore B is the longest. Thus, answer choice B is the best answer.
A 15.0-eV photon collides with and ionizes a hydrogen atom. If the atom was originally in the ground state (ionization potential =13.6 eV), what is the kinetic energy of the ejected electron?
1. 1.4 eV
2. 13.6 eV
3. 15.0 eV
4. 28.6 eV
Answer:

Explanation: Conservation of energy requires that the 15.0 eV photon energy first provides the ionization energy to unbind the electron, and then allows any excess energy to become the electron's kinetic energy. The kinetic energy is this case is 15.0 eV-13.6 eV = 1.4 eV. Thus the correct answer is A.
When 47Be undergoes radioactive decay by electron capture (a form of ß+ decay), the resulting nucleus is:
1. 36Li
2. 37Li
3. 47Be
4. 48Be
Answer: B

Explanation: In radioactive decay, the sum of the mass numbers A and atomic numbers Z, before and after decay, must balance. The numbers for beryllium undergoing positron decay are: mass (7 = 7 + 0) and atomic (4 = 3 + 1). The resulting nucleus is 73Li. Thus, answer choice B is the best answer.
A force F is used to raise a 4-kg mass M from the ground to a height of 5 m

What is the work done by the force F? (Note: sin 60o = 0.87; cos 60o = 0.50. Ignore friction and the weights of the pulleys.)
1. 50 J
2. 100 J
3. 174 J
4. 200 J
Answer: D

Explanation: Work is the product of force and distance. The easiest way to calculate the work in this pulley problem is to multiply the net force on the weight mg by the distance it is raised: 4 kg x 10 m/s2 x 5 m = 200 J. Therefore, answer choice D is the best answer.