Kaplan MCAT General Chemistry Review

Dalton’s law of partial pressures says that a gas’s molar fraction multiplied by the total pressure gives the partial pressure supplied by that specific gas. This question asks specifically about 1.5 mole of NO, out of a total of 3 mole, which means that NO has a molar fraction of X_NO = 0.5. Solve for P using the ideal gas law: (2 L)(P) = (3.0 mol)(R)(300 K), P = 450R atm. This value multiplied by X_NO comes to 225R atm.

43. A

This question simply tests your knowledge of the conditions under which the ideal gas law is most relevant. The reference to experiment 1 is included just to mislead you. It is necessary to know only that the gases act closest to ideal when they are at high temperatures and low pressures.

44. A

At first glance, this question looks extremely simple, yet you must realize that if the center divider is receiving different pressures from each side, it will move until pressure from both sides is equal. When the experimenter reduced the molar concentration in V₂ by half, the pressure was reduced on the V₂ side of the divider, leading to the expansion of V₁. It is important to note that (B) and (D) are incorrect because the question explicitly states that the cylinder is allowed to re-equilibrate with the new molar concentrations.

PASSAGE VI

45. C

An oxidation-reduction reaction requires a transfer of electrons from one atom to another. This reaction simply involves the neutralization of a strong acid by a weak base; no electrons are transferred and all of the oxidation numbers stay constant throughout the reaction.

46. B

The conjugate base of a Brønsted-Lowry acid is the product that does NOT include the H⁺ ion that came from the acid. Even if you did not know this definition, there is a hint to this answer in the passage; because the products in reaction 1 contain a salt and an acid, it is clear that the salt is not the conjugate acid in neutralization reactions. Based on this information, you should be able to determine that the conjugate base is MgCl₂. You can determine the percent composition of the cation (Mg²⁺) in this salt by dividing the molecular weight of the cation (24.3 g/mol) by the molecular weight of the molecule (95.2 g/mol). It should be obvious that 24.3/95.2 is approximately equal to 25/100, which equals 25%.

47. B

The passage states that efficacy is equal to the number of moles of HCl that can be neutralized by one gram of antacid. This means that the most effective antacids are those with the maximum neutralization capacity per gram. Because CO₃^2- can neutralize two H⁺ ions, Al₂(CO₃)₃ has the capacity to neutralize six HCl molecules. Similarly, each of the other answer choices can neutralize three HCl molecules. (C) and (D) can be eliminated because they have the same neutralization capacity as Al(OH)₃, but are significantly heavier molecules. To decide between (A) and (B), consider that Al₂(CO₃)₃ has a molecular weight of 234 g/ mol, while Al(OH)₃ has a molecular weight of 78 g/mol. Because Al₂(CO₃)₃ can neutralize only twice as many molecules as Al(OH)₃ but has nearly three times the weight, it is clear that Al(OH)₃ has the best per-weight efficacy.

48. C

The passage states that NaHCO₃ is an antacid, so (A) is incorrect. The reactions suggested in (C) and (D), where H₂CO₃(aq) decomposes to produce H₂O (l) and CO₂ (g), is a common reaction that recurs every time H₂CO₃ is present in aqueous solution. Because nearly all the CO₂ is in gas form, it does not significantly affect the pH of the solution; this means that there are no acids left to decrease the pH and, therefore, NaHCO₃ is an effective antacid.

49. B

The limiting reagent is the reactant that is completely used up during the reaction while the other reactant still remains. Antacids are alkaline, so if the pH is below 7, there must not have been enough antacid to neutralize all of the HCl. This means that the progress of the reaction was limited by the amount of antacid.

50. A

The passage states that the student initially tested 1 gram of antacid along with 100 mL of 0.1 M HCl. Magnesium hydroxide, Mg(OH)₂, has a molecular weight of about 58 g/mol; because the answer choices are all very rough approximations, we can estimate the weight as about 50 g/mol in order to make the calculations easier. At this molecular weight, 1 gram of antacid is approximately equal to (1 g)/(50 g/mol) = 0.02 mol. 100 mL of 0.1 M HCl is equal to (0.1 L) × (0.1 mol/L) = 0.01 mole. Because each molecule of Mg(OH)₂ has two OH^- ions, only half a mole of Mg(OH)₂ is required to neutralize a mole of acid; therefore, only 0.005 moles of antacid is used up. The student started with 0.02 moles of antacid, so he is now left with 0.02 - 0.005 = 0.015 moles. To convert back to grams, we multiply 0.015 moles by 50 g/mol to get 0.75 g—which is equal to 750 mg.

51. D