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MCAT Content / AAMC MCAT Practice Exam 4 Cp Solutions

AAMC FL4 CP [Web]

Exam 4 C/P Solutions: Passage 1

1) First thing we want to do to answer this question is revisit the part of the passage that talks about limestone. The author tells us in Step 1, “The chemist placed a sample of limestone, CaCO3(s), into a sealed chamber and then heated the limestone to 1200 K to generate CO2(g) and CaO(s).” However, there’s a slight change in the question stem. The limestone, CaCO3(s) is only heated to 900 K, but does not decompose and generate CO2(g) and CaO(s). What does that tell us? The reaction is not spontaneous. Let’s jump into some quick background information we’ll use to answer this question:

ΔG = ΔH – TΔS

We don’t get the decomposition the author talks about in Step 1, and for nonspontaneous reactions, ΔG is positive. Additionally, entropy increases if we go from a solid reactant to a gas in our products, which would mean a positive ΔS. Given this information, ΔG can only be positive in our above equation when ΔH is positive and is greater than TΔS (which we know is also a positive number in this circumstance).

  1. positive and less than TΔS. First part of this answer choice is consistent with our breakdown. We mentioned that for nonspontaneous reactions, ΔG is positive. Entropy increases if we go from a solid reactant to a gas in our products, so we have a positive ΔS, and we’re told temperature is 900 K. Given our equation ΔG = ΔH – TΔS, ΔH must be positive and greater than TΔS in order for ΔG to be positive. That means this answer choice is only half right.
  2. positive and greater than TΔS. This answer choice is entirely consistent with our breakdown of the question. We mentioned some key points in the breakdown that we can reiterate here. For nonspontaneous reactions, ΔG is positive. Entropy increases going from a solid reactant to gas in our products, so we have a positive ΔS. We’re told temperature is 900 K. Given our equation ΔG = ΔH – TΔS, ΔH must be greater than TΔS in order for ΔG to be positive. This is going to be our best answer.
  3. negative and less than TΔS. This answer choice is the opposite of what we said in our breakdown. We expect ΔG to be positive and greater than TΔS.
  4. negative and greater than TΔS. This answer choice is only half correct. We expect ΔG to be positive and greater than TΔS. We can stick with answer choice B as our best answer choice.

2) This question has to do with the same part of the passage we focused on for Question 1. We’re asked about limestone being heated during Step 1.

We’re told an equilibrium is established and we can write out the equation here:

CaCO3 (s) ⇌ CO2 (g) + CaO (s)

The equilibrium constant is the ratio of the mathematical product of the concentrations of the products of a reaction to the mathematical product of the concentrations of the reactants of the reaction. Each concentration is raised to the power of its coefficient in the balanced chemical equation. If we have an equation aA + bB ⇌ cC + dD, we can write out the equilibrium constant expression:

However, before we write anything for this specific question, what we need to pay attention to is the state of each reactant and product. Pure solids and liquids are not included when writing out this equilibrium constant. That means we are not including CaCO3 (s) and CaO (s) from our initial equation. That means we are only left with the product CO2 (g). Our equilibrium constant is simply [CO2].

  1. [CaO] We noted in our equation that CaO is a solid. Pure solids and liquids are not included when writing out our equilibrium constant.
  2. [CaCO3] We noted in our equation that limestone (CaCO3) is a solid. Pure solids and liquids are not included when writing out our equilibrium constant.
  3. [CO2] This answer choice matches our breakdown. This is going to be our best answer choice.
  4. [CaO] × [CaCO3] This answer choice incorrectly includes the solid reactant and product instead of excluding these. We can stick with answer choice C as our correct answer.

3) This is another passage-based question in this set, but once again, the test-maker simplifies our lives a bit by telling us where we will find the necessary information to answer our question. We’re focused on Reaction 2:

Specifically, we want to hone in on nitrogen. The oxidation number describes explicitly the degree to which an element can be oxidized (lose electrons) or reduced (gain electrons). The number is the effective charge on an atom in a compound. We can look at nitrogen in our reactants and products.

First our products: The oxidation number of hydrogen is +1 when it is combined with a nonmetal as in CH4, NH3, H2O, and HCl. In uncharged NH3, each hydrogen has an oxidation number of +1 while nitrogen then has an oxidation number of -3.

In our reactants we have protonation of ammonia so we’re looking at NH4+. Once again, the oxidation number of hydrogen is +1 when it is combined with a nonmetal. In NH4+, each hydrogen has an oxidation number of +1 while nitrogen still has an oxidation number of -3. Together that is where we get the +1 overall on NH4+.

  1. Yes; it changed from –3 to –4. This answer choice correctly lists the initial oxidation number as -3, but we reasoned out that the oxidation number does not change after the protonation of ammonia.
  2. Yes; it changed from 0 to +1. This answer choice is incorrectly telling us the charge on the compounds in which nitrogen is found. That’s not what we’re looking for. We’re looking for the oxidation state of N which remains at -3.
  3. No; it remained at –3. This answer choice matches the breakdown of the question exactly. The oxidation state of N does not change; this is going to be our best answer choice.
  4. No; it remained at +1. This answer choice is describing the oxidation number of hydrogen. We’re looking for the oxidation state of N which remained at -3. Answer choice C is our best answer.

4) This is like a pseudo standalone question. Although the test-maker is asking about something that’s tangentially related to the passage, we’re asked about the approximate volume of a gas at STP. The only thing we’ll need to know from the passage is the number of moles of Gas X.

1 mole of any gas at standard temperature and pressure (273 K and 1 atm) occupies a volume of 22.4 L.

  1. 22.4 L. This answer choice matches what we know from our general knowledge. 1 mole of any gas at standard temperature and pressure (273 K and 1 atm) occupies a volume of 22.4 L.
  2. 44.8 L. This answer choice incorrectly lists the volume for two moles of gas at STP. Answer choice A remains superior.
  3. 67.2 L. This answer choice incorrectly lists the volume for three moles of gas at STP. Answer choice A remains superior
  4. 89.6 L. This answer choice incorrectly lists the volume for four moles of gas at STP. We can eliminate answer choices B, C, and D and keep answer choice A as our correct answer.

 

Exam 4 C/P Solutions: Passage 2

5) This question is asking about a specific aspect of the research conducted in the passage.

We’re focused on Paragraph 3 that discusses the cuvettes being identical in terms of optical properties. Why would the researchers make sure the cuvettes are identical in terms of optical properties? It’s simply good research protocol. The spectrophotometer measures the light that passes through the solution, and having mismatched cuvettes can lead to inconsistent results. The researchers have to ensure the cuvettes have identical optical properties so any differences in absorption can properly be attributed to glucose differences, and not because of cuvette differences.

  1. To enable the comparison of the absorption spectra. This is exactly what we’re looking for in an answer. By ensuring the cuvettes were identical in terms of optical properties, the researchers can ensure they are comparing the absorption spectra and not having their results influenced by differences in cuvettes.
  2. To reduce the absorption in the glass walls. This answer choice is out of scope. Having the glass walls be identical does not mean absorption is reduced. Two identical cuvettes might reduce absorption differently than a different set of identical cuvettes. Ensuring the cuvettes were identical in terms of optical properties does not necessarily mean there is reduced absorption.
  3. To decrease the uncertainty in the wavelength. This answer choice is out of scope. Ensuring the cuvettes have identical optical properties does not decrease uncertainty in wavelength, but rather allows the researchers to focus on differences in absorption.
  4. To increase the absorption in the solutions. Reasoning here is going to be similar to answer choice B. Having the glass walls be identical does not mean absorption increases in the solutions. The solution absorption should not be affected by identical optical properties of the cuvettes. Answer choice A remains our best answer choice.

6) Photon energy can be calculated by using the frequency of the light particles emitted from an object and multiplying by Planck’s constant to work out the energy:

E = hf

Where h is Planck constant. This states that the energy of a photon depends directly on its frequency. By substituting wave properties, this equation can also be written as E = hc/λ which helps use utilize the constants we’re given in this particular question. We’re solving for E, we have a value for hc, and we’re told the wavelength at which absorbance occurred in the passage:

hc = 19.8 × 10–26 J•m and λ=625 nm

Use dimensional analysis to convert wavelength to the proper units:

625 nm x (1 m / 109 nm) = 625 x 10-9 m

E = hc/λ = (19.8 × 10–26 J•m) / (625 x 10-9 m) = approximately 3.2 x 10-19 J.

We can convert Joules to eV using the constant we’re given in the question stem: 1 eV = 1.6 × 10–19 J

Use dimensional analysis again: (3.2 x 10-19 J) x (1 ev / 1.6 × 10–19 J) = 2eV

We did a small bit of approximating, however, a quick glance at our answer choices shows our calculation corresponds to only to answer choice B. Answer choices A, C, and D all list incorrect values.

7) This is another passage-based question in this set that relies on doing some math, the test-maker simplifies our lives by telling us where we will find the necessary information to answer our question. We’re focused on Table 1:

Table 1 Absorbance of Standard Glucose Solutions and of a Diluted Blood Sample

Absorbance Glucose concentration (mg/dL)
0.00 0
0.06 1.5
0.12 3.0
0.18 4.5
0.20 diluted blood sample
0.24 6.0

The passage also tells us there was a 1/30 dilution ratio, so we have to multiply glucose concentration in Table 1 by 30 to find the actual glucose concentration of the glucose in the blood from which the sample was taken. The diluted blood sample is going to be somewhere between 135 mg/dL (corresponding to absorbance of 0.18) and 180 mg/dL (corresponding to absorbance of 0.24). Only answer choice D falls within this range.

Alternatively, we can note the trend in Table 1. Every 0.06 increase in Absorbance means an increase in glucose concentration of 1.5 mg/dL. Therefore, every 0.02 increase in Absorbance should mean an increase in glucose concentration of 0.5 mg/dL. From 0.18 to 0.20 we should see the diluted glucose concentration jump from 4.5 to 5.0 mg/dL. Multiplying 5.0 mg/dL gets us an exact value of 150 mg/dL, or answer choice D.

8) This is a fairly common question type when it comes to experimental passages. We’re asked about modifications to the research we read about in the passage. By doing so, the test-maker ensures that we have grasped the big picture of the experiment and passage. What do I mean by that? If we don’t understand the initial experiment taking place in the passage, it’s very unlikely that we will be able to explain the effects of different modifications.

Both interpolation and extrapolation involve predicting values related to a data set. Interpolation is used to predict values between known data points, while extrapolation is used to predict values outside the range of data. Note the blood sample in the question stem tested above the range of the standards used in the experiment. We want to get that glucose concentration within the range of the standards used in the experiment so the researchers can use data interpolation as opposed to extrapolation.

  1. Increase the enzyme concentration. This is the opposite of what we’re looking for. We can recall from Paragraphs 1 & 2, that by adding more glucose to Reaction 1, we know we have additional chromogen in solution and optical absorbance is affected by this chromogen. This exacerbates our problem in this situation. We already have a blood sample that tested above the range of the standards used in the experiment, and answer choice A would increase this value.
  2. Increase the oxygen pressure. This answer choice is going to be similar to answer choice A. We can look at Reaction 1 and consider the effects of increased oxygen pressure. The reaction will shift to the right so more reactants are produced. Once again, we know we have additional chromogen in solution and optical absorbance is affected by this chromogen. This once again exacerbates our problem in this situation.
  3. Decrease the content of oxygen acceptor. This answer choice is unreasonable. Removing glucose, the oxygen acceptor, would make it so we don’t get any valuable information from the research. Removing glucose makes quantifying the amount of glucose present essentially pointless.
  4. Dilute the sample with additional solvent. This answer choice is consistent with our breakdown of the question. We said the blood sample in the question stem tested above the range of the standards used in the experiment. We want to get that glucose concentration within the range of the standards used in the experiment so the researchers can use data interpolation as opposed to extrapolation. How can we get that glucose concentration within that range? By diluting the sample with additional solvent and decreasing the concentration of glucose. The absorbance will then fall within the range of the standards and we can use data interpolation. Answer choice D is our best answer choice.

 

Exam 4 C/P Solutions: Questions 9-12

9) This is a standalone question that relies on our general knowledge. We can start by going over distillation before jumping into the specific components of the mixture described. Distillation is a purification method where the components of a liquid mixture are vaporized and then condensed and isolated at different boiling points. We’re using fractional distillation here specifically (as opposed to simple distillation). Fractional distillation is used when there is a more complex mixture with not just fixed boiling points, but ranges of temperatures where they boil. The equipment is more complex as well: a fractioning column filled with beads is used to condense different compounds at different temperatures more efficiently. This allows for obtaining greater purity in the end.

1-chlorobutane and 1-butanol are structurally similar, with 1-butanol having a hydroxyl group where 1-chlorobutane has a chloride. Distillation, like we mentioned, takes advantage of different boiling points which would arise from the hydrogen bonding in 1-butanol. For the MCAT, when we see FON (fluorine, oxygen, nitrogen) atoms, we associate with hydrogen bonding which increases boiling point.

  1. Both 1-chlorobutane and 1-butanol are polar. We have to be careful here because while this is a true statement, does it answer the specific question being asked? Separating the molecules in our question stem via fractional distillation relies on boiling point differences and not whether the molecules are polar or nonpolar. Both molecules being polar is actually a shared property of the two molecules and does not explain the boiling point difference.
  2. Both 1-chlorobutane and 1-butanol are nonpolar. Reasoning here is going to be similar to that in answer choice A. This is actually a factually incorrect statement. Both molecules in the question stem are polar. Separating the molecules in our question stem via fractional distillation relies on boiling point differences and not whether the molecules are polar or nonpolar.
  3. The boiling point of 1-chlorobutane is substantially higher than that of 1-butanol. This is another factually incorrect statement. We mentioned in the breakdown of the question, distillation takes advantage of different boiling points that would arise from the hydrogen bonding in 1-butanol. We expect 1-butanol has a higher boiling point.
  4. The boiling point of 1-chlorobutane is substantially lower than that of 1-butanol. This answer choice matches what we said in the breakdown of the question. Distillation takes advantage of different boiling points that would arise from the hydrogen bonding in 1-butanol. We expect 1-butanol has a higher boiling point. Note one of the keys to answering this question was simply knowing background information about distillation. We said distillation is a purification method where the components of a mixture are vaporized, condensed, isolated at different boiling points. By knowing we’re separating based on boiling points we could have narrowed our answer choices down to C and D. We can stick with answer choice D as our correct answer.

10) This is another standalone question in this set that is going to be answered using general knowledge. A quick glance at our answer choices shows we’ll have to know the properties of different functional groups and which are readily oxidized.

A. A primary alcohol to an aldehyde. Primary and secondary alcohols are readily oxidized. We can show the oxidation of ethanol (a primary alcohol) to form acetaldehyde (an aldehyde) below:

B. A tertiary alcohol to a ketone. Tertiary alcohols (R3COH) are resistant to oxidation because the carbon atom that carries the OH group does not have a hydrogen atom attached but is instead bonded to other carbon atoms. The carbon atom bearing the OH group must be able to release one of its attached atoms to form the double bond. The carbon-to-hydrogen bonding is easily broken under oxidative conditions, but carbon-to-carbon bonds are not. Therefore, tertiary alcohols are not easily oxidized.

C. An aldehyde to a carboxylic acid. Due to the free hydrogen seen in aldehydes, aldehydes can be oxidized to transform the carbonyl group (C=O) to a carboxylic acid group (-COOH). Answer choice B remains the best option.

D. A secondary alcohol to a ketone. Primary and secondary alcohols are readily oxidized. We can show the oxidation of isopropyl alcohol to give acetone (the simplest ketone):


Answer choice B remains our best answer choice.

11)

This is a standalone question that relies on our knowledge of naming hydrocarbons according to the IUPAC. First thing we’ll note is the number of carbons in the parent chain. From left to bottom right in the hydrocarbon, we count 5 carbons. Next, we notice a double bond between the 2nd carbon from the left to the 3rd carbon. We want that double bond to have the lowest possible numbers before numbering the methyl group near the top right:

Going by our numbering system above and given the length of the parent chain (5 carbons) we’re looking at 3-methylpent-2-ene. We have to also consider whether the highest priority groups are on the same side or opposite sides of our double bond between carbon 2 and carbon 3. We note they are on the same side, which means we have Z-3-methylpent-2-ene which corresponds to answer choice A. The incorrect options either use the E-designation which would incorrectly imply the highest priority groups are on the opposite side of our double bond between carbon 2 and carbon 3, or they incorrectly count the length of the parent chain as 4 carbons (ethyl). Only answer choice A is the correct systematic name for the hydrocarbon.

12) This is a standalone question that relies on our knowledge of vectors. Vectors are geometric representations of magnitude and direction which are often represented by straight arrows. These arrows start at one point on a coordinate axis and end at a different point. Vectors have a length and direction. Vectors may be added or subtracted graphically by laying them end to end on a set of axes. The graphical method of vector addition is also known as the head-to-tail method. Given the angle in our question stem can vary from 0° to 180°, the vectors can be added head-to-tail to give a resultant vector of magnitudes from 3 units to 13 units. When A and B make an angle of 180° (opposite directions) we have a magnitude of only 3 units. When A and B make an angle of 0° (same direction head-to-tail) we have a magnitude of up to 13 units. The only answer choice that does not fit in this range is answer choice A: 2 units.

 

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