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MCAT Practice Test 6R Solutions ®

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6R

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MCAT

MEDICAL COLLEGE ADMISSION TEST

MCAT P RACTICE T EST 6R S OLUTIONS Edited, produced, typeset, and illustrated by Steven A. Leduc National Director of MCAT Research, Production & Development, The Princeton Review

Jennifer Wooddell Verbal Reasoning Solutions Steven A. Leduc Physics and General Chemistry Solutions Judene Wright Biology Solutions Douglas S. Daniels Organic Chemistry Solutions

Copyright © 2003, 2002 by Princeton Review, Inc. All rights reserved. MCAT is a service mark of the Association of American Medical Colleges (AAMC). TPR is not affiliated with Princeton University or with the AAMC. Version 1.0

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MCAT P RACTICE T EST 6R S OLUTIONS C ONTENTS : Physical Sciences ............. 3 Verbal Reasoning .............. 12

Biological Sciences ....... 30

030316

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PHYSICAL SCIENCES Passage I 1. D. Which of the four compounds listed in the choices would produce a basic (alkaline) solution in a lake? Since KOH (choice D) is a strong base, its dissolution in water would produce a basic solution, so this is the best choice. Choice C, sulfuric acid, would produce an acidic solution, while neither choice A nor B would affect the pH. 2. C. A hydrogen bond is the intermolecular force between the partial positive charge on an H bonded to an F, O, or N atom and the partial negative charge on an F, O, or N atom in another molecule. Methane (CH4) has no F, O, or N atoms, and carbon dioxide (CO2) has no H atoms, so if we’re asked for the pair of compounds that can form extensive networks of hydrogen bonds with both participating, we must eliminate choices A and B (which involve methane) and choice D (which involves carbon dioxide). 3. D. Isotopes of an element differ only in the number of neutrons they contain, not the number of protons; this eliminates choices A and B right away. Carbon-12 has 6 protons and 6 neutrons, while carbon-14 has 6 protons and 8 neutrons. Therefore, these isotopes differ by 8 – 6 = 2 neutrons, choice D. 4. C. Since HCl is a strong acid, it will dissociate completely in solution. So, if [HCl] = 0.01 M, then [H+] = 0.01 M, and pH = –log [H+] = –log (0.01) = –log (10–2) = –(–2) = 2. 5. D. Here are the geometric families and shapes of the given molecule, methane, as well as those of the four choices: H C

H

H H

S H

H

O C O

CH4 geometric family: tetrahedral shape: tetrahedral

H 2S geometric family: tetrahedral shape: bent

F

F Xe F

F

Cl

CO2 geometric family: linear shape: linear

Si

Cl

Cl Cl

XeF4 geometric family: octahedral shape: square planar

SiCl 4 geometric family: tetrahedral shape: tetrahedral

While both H2S and SiCl4 are in the same geometric family as methane, H2S has a bent shape while SiCl4 has a tetrahedral shape. Therefore, of the choices given, SiCl4 (choice D) is most like CH4, and thus the best choice here. 6. B. In a very low pH environment, the amino group of glycine, H2N–CH2–COOH, is protonated (as is the carboxyl group), giving H3N+–CH2–COOH, choice B. 7. A. Helium (choice B) is an inert gas and does not accept electrons. H2 (choice C) is a stable molecule and is therefore unlikely to accept electrons. Iron (choice D) is a metal and is therefore far more likely to donate electrons (becoming a cation) than to accept them. However, sulfur (choice A) is in the oxygen family (“Like oxygen atoms . . .”) and has a high electron affinity, so it’s the best choice here.

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Passage II Note: The questions in this passage are on periodic trends or atomic structure. The passage text is irrelevant. 8. A. Atomic radius generally decreases as we move down a column or across a row to the right in the periodic table. The four atoms listed as choices are all in the same row (Period 3), and sodium (choice A) is farthest to the left, so we would predict that it has the largest radius of the four. 9. D. Ionization energy generally increases as we move up a column or across a row to the right in the periodic table. The four atoms listed as choices are all in the same row (Period 4), and krypton (choice D) is farthest to the right, so we would predict that it has the highest ionization energy of the four. Furthermore, since krypton is a noble gas, it has the particularly stable octet configuration and thus a very high ionization energy. 10. C. Since strontium (Sr) has atomic number 38, a (neutral) strontium atom has 38 electrons. Since the total number of protons and neutrons in strontium-90 is 90, the sum of the numbers of protons, neutrons, and electrons is 90 + 38 = 128. 11. B. Bromine (Br) has atomic number 35, which means that every bromine atom contains 35 protons. If an isotope of bromine has 44 neutrons, then the total number of protons and neutrons—which is the mass number—is 35 + 44 = 79. 12. C. Electronegativity generally increases as we move across a row to the right in the periodic table. As a result, bonds between atoms on the far left with those on the far right (excluding those in Group 18, the noble gases) tend to be highly polar (ionic). Of the four pairs listed as choices, Ca (calcium) and I (iodine) are farthest apart—Ca is in Group 2, I in Group 17—and would therefore be the most likely to form an ionic bond.

Passage III 13. B. Using distance = rate ¥ time, we get d = (3 ¥ 108 m/s) ¥ (7 ¥ 10–2 s) = 21 ¥ 106 m = 2.1 ¥ 107 m. 14. D. Any object moving in a circular path must feel a force that provides the centripetal force (choice D). For a satellite orbiting the earth, it is Earth’s gravitational force that provides the necessary centripetal force. Choice A is wrong because centrifugal force is a fictitious force directed away from the center of the circle and does not keep an object moving in a circular path. Choice B can be eliminated since, for one thing, at the altitude of the GPS satellites, there is no atmosphere. And we can eliminate choice C, since the earth’s gravitational force on the satellite is much stronger than the moon’s (so it doesn’t offset it). 15. A. This is a proportion question using Newton’s Law of Gravitation, F = GMm/r2. If we increase r by a factor of 6 and m by a factor of 4, the ratio m/r2 will change by a factor of 4/62 = 1/9, so F will decrease by a factor of 9: F¢ = G

M (4 m) ( 6r )

2

=

4 Mm 1 Mm 1 ◊ G 2 = ◊G 2 = ◊ F 36 9 9 r r

16. C. The passage tells us that the voltage of the Ni–Cd battery is 1.32 V. Since we know V and we’re given P (the power), we use the equation P = IV to determine the current: I = P/V = (3.96 W)/(1.32 V) = 3 A. 17. C. The last sentence of the passage tells us the two frequencies that are used: f1 = 102.1 MHz and f2 = 104.9 MHz. By definition, the beat frequency is the difference between the two: fbeat = f2 – f1 = 2.8 MHz = 2.8 ¥ 106 Hz. 18. D. The speed of an electromagnetic wave through a medium is given by the equation v = c/n, where n is the medium’s index of refraction. The vacuum of space has n = 1, so v = c; however, through any material medium (such as atmospheric gases), n is greater than 1, so v will be less than c.

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19. A. We’re told in the first line of the second paragraph of the passage that the radioactive cesium source has a mass of 5 grams. So, the question is, How many half-lives must elapse to get this mass down to 0.625 grams? Dividing 5 grams by 2 repeatedly until we reach 0.625 grams, we see that it takes 3 half-lives: 1 half-life

2nd half-life

3rd half-life

æÆ 1. 25 g ææææ æÆ 0. 625 g 5 g æ æææÆ 2. 5 g æ æææ Since each half-life is 175 years, the time period required would be 3 ¥ (175 years) = 525 years.

Passage IV 20. C. The second sentence of the passage states that the “synthesis of Olestra starts with a base-catalyzed cleavage . . .” (emphasis added). Of the choices given, only choice C, NaOH, is a base. 21. C. Since the specific heat of water is 1 cal/g∞C—or, equivalently, 1 kcal/kg∞C—the fact that the temperature of 1 kg of water increased by 50∞C means that the energy added to the water was 50 kcal: Ê 1 kcal ˆ q = mcDT = (1 kg)Á ˜ (50∞C) = 50 kcal Ë kg ◊ ∞C ¯ So, if the combustion of 10 peanuts releases 50 kcal of energy, the combustion of 1 peanut would release 50/10 = 5 kcal. 22. A. In the last sentence of the third paragraph of the passage, we learn that “Olestra is not metabolized. . . .” If it’s not metabolized, it won’t contribute any dietary calories to a human consumer. 23. A. From the structure of glycerine shown in Figure 1 in the passage, we see that this molecule contains 3 –OH groups, while the formula given in the question tells us that isopropyl alcohol contains just 1 –OH group. Therefore, not only does glycerine have a greater molecular weight, but it will have far more hydrogen bonds than isopropyl alcohol. As a result, the boiling point of glycerine should be much higher than that of isopropyl alcohol (choices C and D are eliminated, and A is a better choice than B). [By the way, the BP of glycerine is 290∞C, while the BP of isopropyl alcohol is just 82.5∞C.] 24. B. First, eliminate choices C and D. A mere 20 moles of methanol will not weigh several hundred thousand (and certainly not more than a million) pounds! The molecular weight of methanol, CH3OH, is 12 + 3·1 + 16 + 1 = 32 g/mol. Therefore, 20 moles of CH3OH has a mass of (20 mol)(32 g/mol) = 640 g, or 0.64 kg. Since the last sentence of the passage tells us that 1 kg “equals” 2.2 pounds, we find that 0.64 kg is equivalent to (0.64 kg)(2.2 lb/kg) ª (0.7)(2) = 1.4 lb.

Independent Questions 25. B. Electron affinity generally increases as we move to the right across a row or up within a column in the periodic table. Of the atoms listed as choices, fluorine (choice B) is farthest to the right and highest in the periodic table, so we’d expect that it would have the highest electron affinity. (Furthermore, if F acquires an extra electron, it achieves the much-desired noble gas (octet) configuration, so F “really wants” an extra electron; therefore, it has a high electron affinity.) 26. C. The half-cell reduction reaction of copper is shown directly in the table, Cu+ + e– Æ Cu, with E∞ = +0.52 V. If copper is to be reduced, then the other metal used in the cell must be oxidized, and we must remember that to form a galvanic cell, we need the overall cell voltage to be positive. The oxidation of silver has a potential of –0.80 V (the reverse of the reduction of Ag+ given in the table), so the overall cell voltage here would be –0.28 V, and silver is eliminated from consideration (and so choices A and D are wrong). Note that we don’t need to check lead (because it’s in both of the remaining choices, B and C). Since the oxidation of zinc has a potential of +0.76 V (the reverse of the reduction of Zn2+ given in the table), the overall cell voltage here would be +1.38 V, so zinc will work. Therefore, the answer must be C. 27. A. Frequency ( f, the number of cycles per second) and period (T, the number of seconds per cycle) are reciprocals of each other. Since f = 1/T, we have f T = 1.

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28. A. Use Big Five #2, and call down the positive direction. Then v0 = –v (it’s negative, because v0 is upward and we’re calling down the positive direction) and vfinal = +v, so vfinal = v0 + at becomes v = (–v) + gt, or 2v = gt, which gives t = 2v/g. This means that t is inversely proportional to g. So, if g is decreased by a factor of 6, then t will increase by a factor of 6.

Passage V 29. C. In the reaction 2 Ag + H2S Æ Ag2S + H2, the Ag displaced H from H2S and took its place, forming Ag2S. This is an example of a single replacement (or single displacement) reaction. 30. A. By the stoichiometry of Equation 1 (which is balanced), 2 moles of SO42– ions would produce 2 moles of H2S. Since the molar masses of H and S are 1 and 32, respectively, each mole of H2S has a mass of 2·1 + 32 = 34 g, so the mass of 2 moles of H2S is 2 · 34 g = 68 g. 31. B. Because H2S is a weak acid (as stated in the sentence above Equation 1 in the passage), its ionization constant, Ka, is small. Since most weak acids have Ka values in the range 10–12 to 10–2, we’d certainly expect that the Ka value of H2S is in this range as well, and thus be much less than 1. (Granted it’s small, so we could say that it’s “near zero,” but the fact that Ka values can actually be measured and used in calculations means that we wouldn’t say that all weak acids have ionization constants that are negligible (i.e., near zero); we’d instead simply say that they’re very small. So, A is a good choice, but B is a better choice here.) 32. B. Choices A and D are not reactions between the two compounds; instead, they show how the ions would react to form each compound separately; these choices can therefore be eliminated. Only choices B and C show the ion from one compound reacting with the opposite ion from the other compound. Choice B shows the reaction that forms “the insoluble CaCO3, which crystallize[s], encapsulating . . . coins, sand, and decaying matter into rock-like clumps . . . [which is what] the explorers who discovered the treasure [in the ocean near Cuba] found . . . ,” as mentioned in the passage. This is the best choice here. Choice C shows the formation of NaCl(s), which we know is soluble, especially since the passage talks about finding treasure in the ocean, where NaCl would be dissolved into its separate ions. 33. D. Let’s see how the oxidation numbers of the atoms change as a result of the reaction: oxidation numbers:

0

+1 –2

+1 –2

0

2 Ag(s) + H2S(aq) Æ Ag2S(s) + H2(g)

Since Ag was oxidized (its oxidation number increased), it acted as the reducing agent.

Passage VI 34. D. We learn in the second-to-last sentence of the first paragraph of the passage that tones separated by a perfect fifth differ in frequency by a factor of 2/3. This means that the higher-frequency tone has a frequency that is 3/2 times the lowerfrequency tone when the tones are separated by a perfect fifth. Since the equation given in the passage tells us that f is proportional to T1/2 (where T is the tension in the string), if we want to increase f by a factor of 3/2, then we have to increase T by a factor of (3/2)2, which is 9/4. 35. C. According to the equation given in the passage, f = (T/r )1/2/(2L), we can lower f by lowering the string’s tension T, increasing the string’s linear mass density r , or increasing the string’s length L. Of the choices given, only C (using heavier strings—that is, strings with a greater linear mass density r) corresponds to one of these options. 36. D. In the last paragraph, the passage states that “the best violins produce loud tones over the full frequency range of the instrument . . .” (emphasis added). Each string has a characteristic fundamental frequency, so in order to produce loud tones over the instrument’s full frequency range, we’d want the violin to have good resonance at the fundamental frequencies of all the strings.

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37. B. In the second paragraph of the passage, we learn that the A string of the violin or viola has a frequency of 440 Hz, and that “the cello is tuned one octave below the viola, which means the frequencies of the cello strings are half that of the viola strings.” So, if the A string of the viola is tuned to 440 Hz, the A string of the cello is tuned to 21(440 Hz) = 220 Hz. 38. D. In the second paragraph of the passage, we learn that the A string of the violin has a frequency of 440 Hz, so a list of the four fundamental frequencies of the violin must include 440 Hz. This eliminates choices A and B. We also learn that the violin’s strings are tuned with decreasing frequency to the notes E, A, D, G. So, if A is 440 Hz, then E must be greater than 440 Hz. Only choice D includes a frequency higher than 440 Hz.

Passage VII 39. B. The atom is attached to 2 springs, each of force constant K. When an atom displaced a distance x from its equilibrium position, each spring exerts a force of magnitude K x , so the magnitude of the total force on the atom is 2K x . This means that the effective spring constant for this system is 2K. 40. D. Frequency ( f, the number of cycles per second) and period (T, the number of seconds per cycle) are reciprocals—or inverses—of each other. 41. C. The question tells us to use a “dimensional argument” to identify the correct formula for T in terms of K and M. This means that all we need to check is that the dimensions (or units) of the two sides of the equation match. The left-hand side of each equation in the answer choices has units of seconds2, since the period T is measured in seconds and p is a constant. Therefore, we need to determine which right-hand side also has units of seconds2. From the equation F = Kx , we know that the units of K are N/m, which is the same as kg/s2. So, to get s2 out of this, we need to flip K—to get s2/kg— and then multiply by kg (mass). Therefore, the right-hand side of the equation must contain the term (1/K)·M, or M/K. Only choice C contains this term. 42. A. For an oscillatory system without dissipative effects due to friction, the total mechanical energy, E, is conserved. As the oscillator moves, there is a continuous trade-off between kinetic and potential energy. The kinetic energy and the potential change individually, but their sum, E = KE + PE, remains constant (choice A). Note that momentum is constantly changing since the speed and direction of the oscillator are constantly changing. 43. D. In the first line of the second paragraph of the passage, we’re given the formula for the potential energy of a spring: PE = Kx2/2. Since there are 2 springs here, the potential energy will be Kx2/2 + Kx2/2 = Kx2. When the atom is at its greatest distance from equilibrium—that is, when x = ±A, where A is the amplitude—there is no kinetic energy, only potential energy. Therefore, E = KE + PE = 0 + KA2 = KA2. Solving the equation E = KA2 for K, we get K = E/A2 (choice D). We could have also answered this question by just using a dimensional argument, like in Question 41 above. Since [E] = J = N·m, and [K] = N/m, we’d need to divide N·m by m2 to get N/m. That is, we’d have to divide E by A2 to get K, so the answer must be D. 44. B. From the equation q = mcDT, where q is heat (energy), m is mass, c is specific heat, and DT is the temperature change, we find that c = q/(mDT), so [c ] =

J [q ] = [m ][DT ] kg ◊ K

45. A. The first sentences of the passage say, “let us assume that each atom can oscillate about its equilibrium position. Interactions with neighboring atoms hold it in place. . . .” Therefore, the model described is limited by this assumption, which holds very well for solids. However, we know that the particles of a monatomic gas move very freely; they are not restricted to small oscillations around fixed equilibrium positions established by interactions with neighboring particles. (In fact, in the kinetic–molecular theory of ideal gases, we assume that there are no such interactions.) Since the basic assumption of the model fails for gases, it can’t be used to describe gases.

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Passage VIII 46. B. Compound 1 contains a four-membered ring, while Compound 2 contains a six-membered ring, so their structural formulas are clearly different. Also, Compound 1 contains 2 B’s, 2 P’s, 2 H’s, and 2 NR2 groups, while Compound 2 contains 3 B’s, 3 P’s, 3 H’s, and 3 NR2 groups, so their molecular formulas are different, too: B2P2H2(NR2)2 vs. B3P3H3(NR2)3. However, their empirical formulas—which give the smallest whole numbers that specify the correct ratios of the atoms—are the same: BPHNR2. 47. C. From the stoichiometry of Equation 1a (which we assume is balanced), the given initial amounts of reactants will produce 0.003 mol of PH3(g). Since the volume of 1 mole of gas...