AAMC FL 4 EXAM QUESTIONS WITH CORRECT ANSWERS LATEST UPDATE 2025-2026
Limestone does NOT decompose when heated to 900 K because, at 900 K, ΔH is:
A. Positive and less than TΔS
B. Positive and greater than TΔS
C. Negative and less than TΔS
D. Negative and greater than TΔS - Answers B.
The reaction does not occur (is not spontaneous). This indicates that the ΔG =ΔH-TΔS >0. From
inspection of the reaction, it can be concluded that ΔS>0. Consequently, ΔH>TΔS explains why
the reaction does not occur
When limestone is heated during Step 1, an equilibrium is established. Which of the following
expressions is the equilibrium constant for the decomposition of limestone?
A. [CaO]
B. [CaCO3]
C. [CO2]
D. [CaO] x [CaCO3] - Answers C.
From the law of mass action, an equilibrium constant expression involves a ration of products
to reactants with exponents determined from the stoichiometry of the reaction. Furthermore,
solids are excluded from equilibrium constant expressions. CO2 (g), as the only non-solid
material in the reaction, is the only substance that appears in the equilibrium constant
expression.
During Reaction 2, did the oxidation state of N change?
A. Yes; it changed from -3 to -4
B. Yes; it changed from 0 to +1
C. No; it remained at -3
D. No; it remained at +1 - Answers C.
The part of the Reaction 2 that involves nitrogen is the protonation of ammonia (NH3 + H-->
NH4+). Acid-base reactions doe not involve oxidation state changes. Furthermore the oxidation
,states of N in NH3 is -3 not 0.
If all of Gas X (from Step 6) is held in a sealed chamber at STP, what will be its appropriate
volume?
A. 22.4 L
B. 44.8 L
C. 67.2 L
D. 89.6 L - Answers A.
The quantity of Gas X was given as 1 mole. One mole of gas occupies 22.4 L at STP.
Why was it important that the cuvettes containing the glucose oxidase and the blood sample
were identical in terms of optical properties?
A. To enable the comparison of the absorption spectra
B. To reduce the absorption in the glass walls
C. To decrease the uncertainty in the wavelength
D. To increase the absorption in the solutions - Answers A.
The identical optical properties of the cuvettes ensure that the absorbed radiation is due only to
the presence of glucose in the blood and not due to the difference in the absorption features of
the walls.
What is the approximate energy of a photon in the absorbed radiation that yielded the data in
Table 1?
A. 1 eV
B. 2 eV
C. 3 eV
D. 4 eV - Answers B. ( I chose C)
,The photon energy is E=hc/λ = 19.8 x 10^-26 J.m/ (625 x 10^-9) = 3.1 x10^-19 J , so about 2 eV
According to Table 1, what is the concentration of the glucose in the blood from which the
diluted sample was taken?
A. 60 mg/dL
B. 90 mg/dL
C. 120 mg/dL
D. 150 mg/dL - Answers D.
From Table 1, the glucose concentration in the diluted sample is (o.20/0.24) x 6.0 mg/dL = 5.0
mg/dL.
The blood then has a glucose concentration of 30 x 5.0 mg/dL= 150 mg/dL.
Suppose a blood sample tested above the range (6.0 mg/dL) of the standards used in the
experiment. What modification will provide a more precise reading by data interpolation as
opposed to extrapolation using the same standards?
A. Increase the enzyme concentration.
B. Increase the oxygen pressure.
C. Decrease the content of the oxygen acceptor
D. Dilute the sample with additional solvent. - Answers D.
By adding solvent, the concentration of glucose will be lowered, and the resulting absorbance
will fall within the range of the standards. This is easily accomplished, and the resulting
calculations that account for the dilution are not difficult.
Which of the following reasons best explains why it is possible to separate a 1:1 mixture of 1-
chlorobutane and 1-butanol with fractional distillation?
A. Both 1-chlorobutane and 1-butanol are polar
, B. Both 1-chlorobutane and 1-butanol are nonpolar
C. The boiling point of 1-chlorobutane is substantially higher than that of 1-butanol
D. The boiling point of 1-chlorobutane is substantially lower than that of 1-butanol - Answers D.
The fact that 1-chlorobutane will have a boiling point that is substantially lower than that of 1-
butanol can be rationalized from chemical principles. The molecules have similar molecular
weights, but 1-butanol has a hydroxyl functional group that can participated in hydrogen
bonding. Hydrogen bonding is a particularly strong force of intermolecular attraction.
Which of the following oxidative transformations is unlikely to occur?
A. A primary alcohol to an aldehyde
B. A tertiary alcohol to a ketone
C. An aldehyde to a carboxylic acid
D. A secondary alcohol to a ketone - Answers B.
Oxidation of tertiary alcohols is difficult because it involved C-C bond breaking
According to the IUPAC, which is the systematic name for the hydrocarbon shown?
A. Z-3-methylpent-2-ene
B. E-3-methylpent-2-ene
C. Z-3-ethylbut-2-ene
D. E-3-ethylbut-2-ene - Answers A.
By IUPAC rules, first identify the longest unbroken chain of carbon atoms. Next, number the
carbon atoms in this chain starting from the end that gives the C=C the lowest numbers. The
double bond is identified by the portion of the carbon atoms from the lowest numbers ed (2),
and then the methyl group is assigned at the 3-position. The stereochemical designator for the
double bond is X because the highest priority groups occur on the same side of the double bond.
(On Z same side)
Two vector magnitudes |A| = 8 units and |B| =5 units make an angle that can vary from 0° to
Limestone does NOT decompose when heated to 900 K because, at 900 K, ΔH is:
A. Positive and less than TΔS
B. Positive and greater than TΔS
C. Negative and less than TΔS
D. Negative and greater than TΔS - Answers B.
The reaction does not occur (is not spontaneous). This indicates that the ΔG =ΔH-TΔS >0. From
inspection of the reaction, it can be concluded that ΔS>0. Consequently, ΔH>TΔS explains why
the reaction does not occur
When limestone is heated during Step 1, an equilibrium is established. Which of the following
expressions is the equilibrium constant for the decomposition of limestone?
A. [CaO]
B. [CaCO3]
C. [CO2]
D. [CaO] x [CaCO3] - Answers C.
From the law of mass action, an equilibrium constant expression involves a ration of products
to reactants with exponents determined from the stoichiometry of the reaction. Furthermore,
solids are excluded from equilibrium constant expressions. CO2 (g), as the only non-solid
material in the reaction, is the only substance that appears in the equilibrium constant
expression.
During Reaction 2, did the oxidation state of N change?
A. Yes; it changed from -3 to -4
B. Yes; it changed from 0 to +1
C. No; it remained at -3
D. No; it remained at +1 - Answers C.
The part of the Reaction 2 that involves nitrogen is the protonation of ammonia (NH3 + H-->
NH4+). Acid-base reactions doe not involve oxidation state changes. Furthermore the oxidation
,states of N in NH3 is -3 not 0.
If all of Gas X (from Step 6) is held in a sealed chamber at STP, what will be its appropriate
volume?
A. 22.4 L
B. 44.8 L
C. 67.2 L
D. 89.6 L - Answers A.
The quantity of Gas X was given as 1 mole. One mole of gas occupies 22.4 L at STP.
Why was it important that the cuvettes containing the glucose oxidase and the blood sample
were identical in terms of optical properties?
A. To enable the comparison of the absorption spectra
B. To reduce the absorption in the glass walls
C. To decrease the uncertainty in the wavelength
D. To increase the absorption in the solutions - Answers A.
The identical optical properties of the cuvettes ensure that the absorbed radiation is due only to
the presence of glucose in the blood and not due to the difference in the absorption features of
the walls.
What is the approximate energy of a photon in the absorbed radiation that yielded the data in
Table 1?
A. 1 eV
B. 2 eV
C. 3 eV
D. 4 eV - Answers B. ( I chose C)
,The photon energy is E=hc/λ = 19.8 x 10^-26 J.m/ (625 x 10^-9) = 3.1 x10^-19 J , so about 2 eV
According to Table 1, what is the concentration of the glucose in the blood from which the
diluted sample was taken?
A. 60 mg/dL
B. 90 mg/dL
C. 120 mg/dL
D. 150 mg/dL - Answers D.
From Table 1, the glucose concentration in the diluted sample is (o.20/0.24) x 6.0 mg/dL = 5.0
mg/dL.
The blood then has a glucose concentration of 30 x 5.0 mg/dL= 150 mg/dL.
Suppose a blood sample tested above the range (6.0 mg/dL) of the standards used in the
experiment. What modification will provide a more precise reading by data interpolation as
opposed to extrapolation using the same standards?
A. Increase the enzyme concentration.
B. Increase the oxygen pressure.
C. Decrease the content of the oxygen acceptor
D. Dilute the sample with additional solvent. - Answers D.
By adding solvent, the concentration of glucose will be lowered, and the resulting absorbance
will fall within the range of the standards. This is easily accomplished, and the resulting
calculations that account for the dilution are not difficult.
Which of the following reasons best explains why it is possible to separate a 1:1 mixture of 1-
chlorobutane and 1-butanol with fractional distillation?
A. Both 1-chlorobutane and 1-butanol are polar
, B. Both 1-chlorobutane and 1-butanol are nonpolar
C. The boiling point of 1-chlorobutane is substantially higher than that of 1-butanol
D. The boiling point of 1-chlorobutane is substantially lower than that of 1-butanol - Answers D.
The fact that 1-chlorobutane will have a boiling point that is substantially lower than that of 1-
butanol can be rationalized from chemical principles. The molecules have similar molecular
weights, but 1-butanol has a hydroxyl functional group that can participated in hydrogen
bonding. Hydrogen bonding is a particularly strong force of intermolecular attraction.
Which of the following oxidative transformations is unlikely to occur?
A. A primary alcohol to an aldehyde
B. A tertiary alcohol to a ketone
C. An aldehyde to a carboxylic acid
D. A secondary alcohol to a ketone - Answers B.
Oxidation of tertiary alcohols is difficult because it involved C-C bond breaking
According to the IUPAC, which is the systematic name for the hydrocarbon shown?
A. Z-3-methylpent-2-ene
B. E-3-methylpent-2-ene
C. Z-3-ethylbut-2-ene
D. E-3-ethylbut-2-ene - Answers A.
By IUPAC rules, first identify the longest unbroken chain of carbon atoms. Next, number the
carbon atoms in this chain starting from the end that gives the C=C the lowest numbers. The
double bond is identified by the portion of the carbon atoms from the lowest numbers ed (2),
and then the methyl group is assigned at the 3-position. The stereochemical designator for the
double bond is X because the highest priority groups occur on the same side of the double bond.
(On Z same side)
Two vector magnitudes |A| = 8 units and |B| =5 units make an angle that can vary from 0° to
