MCAT General Chemistry Review - Alexander Stone Macnow, MD 2019-2020
Solutions
Explanations to Discrete Practice Questions
1. A
The equation ΔT_{b} = iK_{b}m can be used to solve this problem. The change in boiling point is 101.0 − 100 = 1.0°C. Then, plug that into:
The van’t Hoff factor for this solute is 1 because the molecule does not dissociate into smaller components. Then, convert to grams of solute using the definition of molality:
The mass used in this equation is 0.1 kg because 100 mL of water has a mass of 0.1 kg. Then, determine the molar mass: which is closest to (A).
2. DAll three choices can make a solution as long as the two components create a mixture that is of uniform appearance (homogeneous). Hydrogen in platinum is an example of a gas in a solid. Brass and steel are examples of homogeneous mixtures of solids. The air we breathe is an example of a homogeneous mixture of gases; while these are more commonly simply referred to as mixtures, they still fit the criteria of a solution.
3. BBenzene and toluene are both organic liquids and have very similar properties. They are both nonpolar and are almost exactly the same size. Raoult’s law states that ideal solution behavior is observed when solute—solute, solvent—solvent, and solute—solvent interactions are all very similar. Therefore, benzene and toluene in solution will be predicted to behave as a nearly ideal solution.
4. CMelting point depresses upon solute addition, making (A) and (B) incorrect. Solute particles interfere with lattice formation, the highly organized state in which solid molecules align themselves. Colder-than-normal conditions are necessary to create the solid structure.
5. DThe first step will most likely be endothermic because energy is required to break molecules apart. The second step is also endothermic because the intermolecular forces in the solvent must be overcome to allow incorporation of solute particles. The third step will most likely be exothermic because polar water molecules will interact with the dissolved ions, creating a stable solution and releasing energy.
6. CCaS will cause the most negative contribution to ΔS°_{soln} through hydration effects because the Ca^{2+} and S^{2−} ions have the highest charge density compared to the other ions. All of the other ions have charges of +1 or —1,whereas Ca^{2+} and S^{2−} each have charges with a magnitude of 2.
7. BFormation of complex ions between silver ions and ammonia will cause more molecules of solid AgCl to dissociate. The equilibrium is driven toward dissociation because the Ag^{+} ions are essentially being removed from solution when they complex with ammonia. This rationale is based upon Le Châtelier’s principle, stating that when a chemical equilibrium experiences a change in concentration, the system will shift to counteract that change.
8. BThe mass percent of a solute equals the mass of the solute divided by the mass of the total solution times 100%.
Plug in the values given for sucrose, the volume of water and the density of water to determine the %mass of sucrose.
Keep in mind that in rounding while calculating, the denominator was estimated to be larger than the actual value, thus giving an answer that is slightly lower than the actual value. Thus the correct answer is (B), 25.5%. (A) results if rounding error is not taken into account. While these answers are very close, the mass of the water must be slightly less than 300 g, given the density value, so the percent composition of sucrose must be slightly higher than 25%. If the solute’s mass is not added to the solvent’s, the calculated value is 34.2%, which is (D). (C) neglects both the addition step and the rounding error.
9. AMixtures that have a higher vapor pressure than predicted by Raoult’s law have stronger solvent—solvent and solute—solute interactions than solvent—solute interactions. Therefore, particles do not want to stay in solution and more readily evaporate, creating a higher vapor pressure than an ideal solution. Two liquids that have different properties, like hexane (hydrophobic) and ethanol (hydrophilic, small) in (A), would not have many interactions with each other and would cause positive deviation; i.e. higher vapor pressure. (B) and (C) are composed of liquids that are similar to one another and would not show significant deviation from Raoult’s law. (D) contains two liquids that would interact very well with each other, which would actually cause a negative deviation from Raoult’s law—when attracted to one other, solutes and solvents prefer to stay in liquid form and have a lower vapor pressure than predicted by Raoult’s law.
10. ADissolution is governed by enthalpy and entropy, which are related by the equation ΔG°_{soln} = ΔH°_{soln} − TΔS°_{soln}. The cooling of the solution indicates that heat is used up in this bond-breaking reaction. In other words, dissolution is endothermic, and ΔH is positive. The reaction is occurring spontaneously, so ΔG must be negative. The only way that a positive ΔH can result in a negative ΔG is if entropy, ΔS, is a large, positive value as in (A). Conceptually, that means that the only way the solid can dissolve is if the increase in entropy is great enough to overcome the increase in enthalpy. (B) is incorrect because it is clearly stated in the question stem that KCl dissolves; further, all salts of Group 1 metals are soluble. (C) is incorrect because ΔS°_{soln} must be positive in order for KCl to dissolve. Finally, (D) is incorrect because solute dissolution would cause the boiling point to elevate, not depress. It is also not a piece of evidence that could be found simply by observing the beaker’s temperature change.
11. B
The equation to determine the change in boiling point of a solution is as follows: ΔT_{b} = iK_{b}m. m is the molality of the solution, and K_{b} is the boiling point elevation constant. In this case, the solvent is always water, so K_{b} will be the same for each solution. What is needed is the number of dissociated particles from each of the original species. This is referred to as the van’t Hoff factor (i) and is multiplied by molality to give a normality (the concentration of the species of interest—in this case, all particles). The normality values determine which species causes the greatest change in boiling point.
Species |
Number of Moles |
Number of Dissolved Particles |
i × m (Normality) |
CaSO_{4} |
0.4 |
2 |
0.8 |
Fe(NO_{3})_{3} |
0.5 |
4 |
2.0 |
CH_{3}COOH |
1.0 |
Between 1 and 2 (acetic acid is a weak acid and a low percentage of the molecules will dissociate into 2 particles) |
Between 1.0 and 2.0 |
C_{12}H_{22}O_{11} |
1.0 |
1 |
1.0 |
The choice is between iron(III) nitrate and acetic acid. The fact that acetic acid is a weak acid indicates that only a few particles will dissociate into H^{+} and acetate. Therefore, the normality of the acetic acid will be much closer to 1.0 than 2.0.
12. BOsmotic pressure is given by the formula Entering the values from the question stem gives:
Notice that the concentration of seawater is given for all solutes, which represents i×M. It is also given in mOsm/L, which is converted to moles per liter by multiplying by 10^{-3}. Also, the question asks for the minimum pressure required, which means that the correct answer choice must be slightly above the calculated pressure in order for reverse osmosis to proceed.
13. B
200 ppb of Pb^{2+} is equivalent to 200 grams of Pb^{2+} in 10^{9} grams of solution; given the extremely low concentration of lead, the mass of the water can be assumed to be approximately 10^{9} grams, as well. To solve, set up a dimensional analysis question. The units needed at the end are moles per liter (molarity), so convert from grams of lead to moles of lead and grams of water to liters of water:
Note that the denominator was rounded to a smaller number, meaning the estimated answer is slightly larger than the actual value.
14. DSince both salts have a formula MX_{3} (one of one particle, three of another), it is possible to directly compare the molar solubilities of each. When the solutions are mixed, [OH^{−}] is above saturation levels for both the cobalt and the thallium in the solution. Since thallium(III) hydroxide has a smaller K_{sp} than that of cobalt(III) hydroxide, it will react first. The ion product of the mixed solution is higher than the K_{sp} for thallium(III) hydroxide, and the system will shift left to precipitate solid thallium(III) hydroxide. After the thallium(III) hydroxide precipitates, a small excess of OH^{-} will remain, which gives an ion product slightly above the K_{sp} of cobalt (III) hydroxide. This will cause a small amount (1%—3%) of cobalt(III) hydroxide to also precipitate.
15. D
The solubility of AgBr can be determined using the K_{sp} value given in the equation. Some amount of AgBr will dissolve; this is the molar solubility x for these conditions. When AgBr dissociates, there will be x amount of silver(I) formed and x amount of bromide—which is added to the 0.0010 M Br^{-} already present from NaBr.
Given the K_{sp} of 5.4 × 10^{—13}, x will be negligible compared to 0.0010 M. Thus, the math can be simplified to:
Therefore, x, the molar solubility, is 5.35 × 10^{−10}, which looks like (C). However, the units of the answer choices are grams per liter, not molarity, and the result must be multiplied by the molar mass
which is close to (D). Note that a very accurate approximation was reached by rounding down the first number and rounding up the second, balancing the error.