Kaplan MCAT General Chemistry Review

The equation T_b = iK_bm can be used to solve this problem. The change in boiling point is found by subtracting the boiling point of water (the solvent), 100°C, from the elevated boiling point, 101.11°C. Using the given value for K_b, we solve for the molality of the solution and get 2.17 moles/kg. Convert to grams by dividing by 1,000 and then multiply by the mass of the solution, 100 g, to get the moles of solute. To obtain the molar mass, divide the mass of the solute, 70 g, by the number of moles, 0.217 moles, to get a molar mass of 322.58 g/mole. (B) would indicate a problem in unit conversions because the answer is off by an order of magnitude. (C) and (D) are wrong and would result from multiplying 1.11 by the K_b instead of dividing. (Note: Sugar does not dissociate in water, thus the van’t Hoff factor (i) is equal to one.)

2. D

All three choices can make a solution as long as the two components create a mixture that is of uniform appearance (homogenous). Hydrogen in platinum is an example of a gas in a solid. The air we breathe is an example of a homogenous mixture of a gas in a gas. Brass and steel are examples of homogenous mixtures of solids.

3. B

Benzene 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 very similar. Therefore, benzene and toluene in solution will be predicted to behave as a nearly ideal solution. (A) states that the liquids would follow Raoult’s law because they are different. It is true that the compounds are slightly different, but the difference is negligible in terms of Raoult’s law. (C) and (D) are incorrect because they state that the solution would not obey Raoult’s law.

4. A

If the membrane became permeable to water, then water molecules would move to the side with the higher solute concentration, according to the principles of osmosis. The left side has a higher concentration of solute (NaCl), so water will move toward the left in an attempt to balance the concentrations of each side. The level on the left will rise because of the excess water molecules, and the level on the right will fall due to a loss of water molecules.

5. D

The 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 and release energy.

6. C

CaS will cause the most negative S°_soln because the Ca²⁺ 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 Ca²⁺ and S^2- each have charges with an absolute value of 2. To arrange all four species in order of highest to lowest charge density, we’d have to take ion size into account. Smaller ions have higher charge densities. For example, LiF will have a higher charge density than KCl. It follows that the S°_soln is more negative for LiF than for KCl.

7. D

A nonelectrolyte solution will not dissociate into ions in solution. Its effective molarity in solution will be the same as the number of moles that were dissolved. On the other hand, an electrolyte like Mg(NO₃)₂ will dissociate into three ions (Mg and 2NO₃^-). The effective molarity, which is important for colligative properties, will be three times the number of moles that were dissolved. Osmotic pressure is a colligative property and will, therefore, be three times larger for Mg(NO₃)₂ compared to a nonelectrolyte. The molarity of Mg(NO₃)₂, 0.02 M, is two times larger than the nonelectrolyte solution (0.01 M). The nonelectrolyte’s osmotic pressure will have to be multiplied by three and then by two for Mg(NO₃)₂: 15 mm Hg × 3 × 2 equals 90 mm Hg. We can use the osmotic pressure formula to check our work.

Osmotic pressure: = MRT

R terms cancel, and we’re solving for ₂, so we’re left with

8. B

The mass percent of a solute equals the mass of the solute divided by the mass of the total solution. To find the mass of the solution, first find the mass of the solvent, water. Multiplying the volume of the solution by the density gives a mass of 292.5 grams of water. Adding 100 grams of sugar yields a solution with a mass of 392.5 grams. Next, divide 100 grams of sugar by 392.5 grams and multiply by 100 to get a percentage. (A) can be arrived at if water’s density at 80°C is assumed to be 1 g/mL. If we had forgotten to add the solute’s mass to the solvent’s, we’d have calculated 34.2 (100/292.5) percent, which is (D). (C) neglects both the addition step and the correct density value (100/300 = 33.3%).

9. A