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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 (20o 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) pKa
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:

  1. 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.

  2. 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

  3. 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 pKa 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.

  4. 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+] = Ka and pH = pKa. 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.

  5. 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 (20oC).” Table 1 shows that the melting point of crotonic acid is 71.6oC, which means it is a solid at room temperature (i.e., it melts 51.6oC above room temperature). Thus, answer choice C is the best answer.

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