SAT Subject Test Chemistry
REVIEW OF MAJOR TOPICS
Liquids, Solids, and Phase Changes
1. Distillation of water cannot remove
(A) volatile liquids like alcohol
(B) dissolved salts
2. The ratio in water of hydrogen to oxygen by mass is
(A) 1 : 9
(B) 2 : 1
(C) 1 : 2
(D) 1 : 8
3. Decomposing water by an electric current will give a ratio of hydrogen to oxygen by volume equal to
(A) 1 : 9
(B) 2 : 1
(C) 1 : 2
(D) 1 : 8
4. If 10.0 grams of ice melts at 0°C, the total quantity of heat absorbed is
(A) 10.0 J
(B) 334 J
(C) 3,340 J
(D) 33,400 J
5. To heat 10.0 grams of water from 4°C to 14°C requires
(A) 10.0 J
(B) 4.18 J
(C) 418. J
(D) 4180. J
6. The abnormally high boiling point of water in comparison to similar compounds is due primarily to
(A) van der Waals forces
(B) polar covalent bonding
(C) dipole insulation
(D) hydrogen bonding
7. A metallic oxide placed in water would most likely yield
(A) an acid
(B) a base
(C) a metallic anhydride
(D) a basic anhydride
8. A solution can be both
(A) dilute and concentrated
(B) saturated and dilute
(C) saturated and unsaturated
(D) supersaturated and saturated
9. The solubility of a solute must indicate
(A) the temperature of the solution
(B) the quantity of solute
(C) the quantity of solvent
(D) all of the above.
10. A foam is an example of
(A) a gas dispersed in a liquid
(B) a liquid dispersed in a gas
(C) a solid dispersed in a liquid
(D) a liquid dispersed in a liquid
11. When another crystal was added to a water solution of the same substance, the crystal seemed to remain unchanged. Its particles were
(A) going into an unsaturated solution
(B) exchanging places with others in the solution
(C) causing the solution to become supersaturated
(D) not going into solution in this static condition
12. A 10.% solution of NaCl means that in 100. grams of solution there is
(A) 5.85 g NaCl
(B) 58.5 g NaCl
(C) 10. g NaCl
(D) 94 g of H2O
13. The molar mass of AlCl3 is
(A) 62.5 g/mol
(B) 67.9 g/mol
(C) 106.5 g/mol
(D) 133.5 g/mol
14. The molarity of a solution made by placing 98 grams of H2SO4 in sufficient water to make 500. milliliters of solution is
15. If 684 grams of sucrose (molecular mass = 342 g/mol) is dissolved in 2,000 grams of water (essentially 2 L), what will be the freezing point of this solution?
The following questions are in the format that is used on the SAT Subject Test in Chemistry. If you are not familiar with these types of questions, study before doing the remainder of the review questions.
Directions: Each of the following sets of lettered choices refers to the numbered questions immediately below it. For each numbered item, choose the one lettered choice that fits it best. Every choice in a set may be used once, more than once, or not at all.
Questions 16–18 refer to the following graphs:
16. Which of the graphs represents the solubility curve of a substance that has no change in its solubility as the temperature increases?
17. Which of the graphs represents the amount of a solute in a solution as the solute is added even after the saturation point is reached and no more can go into solution?
18. Which of the graphs represents the temperature of a sample of ice over time as it is heated at an even rate from below its freezing point to a temperature where it is in the liquid state above the freezing point?
(A) 1 molar
(B) 1 molal
(C) .5 molar
(D) .5 molal
(E) 25 molar
19. The concentration of a solution of Ca(OH)2 when 74 grams are completely dissolved in a container holding 2 liters of water?
20. The concentration of a solution of sucrose, C12H22O11, when 684 grams are completely dissolved in 2,000 grams of water?
Answers and Explanations
1. (A) Volatile liquids cannot be removed by distillation because they would come over with the steam in this process.
2. (D) In the molar mass of water, there are 2 hydrogen = 2 g and 1 oxygen = 16 g. So the H/O ratio by mass is
2 g/16 g or 1/8.
3. (B) In the balanced reaction, 2H2O → 2H2(g) + O2(g), 2 volumes of hydrogen are released to 1 volume of oxygen, so the ratio is 2:1.
4. (C) It takes 334 J to melt 1 g (heat of fusion for 1 gram) so 10 g × 334 J/g = 3,340 J to melt the ice.
5. (C) m (mass) × ΔT (change of temperature) × specific heat = q (quantity of heat).
10.0 g × (14° - 4°= 10°) × 4.18 J/°/g = 418. J.
6. (D) Because hydrogen bonding is so strong in water, it takes more heat energy to reach its boiling point.
7. (B) Metal oxides are acid anhydrides, because they react with water to form acids. Example:
CaO + H2O → Ca(OH)2
8. (B) Saturated means that the solution is holding all the solute it can at that temperature. Dilute means that there is a small amount of solute in solution compared with the amount of solvent. So a substance that is only slightly soluble can form a saturated dilute solution.
9. (D) Solubility of a solute must give the amount of solute dissolved in a given amount of solvent at a particular temperature.
10. (A) A foam is an example of a colloidal dispersion of a gas in a liquid.
11. (B) This situation describes a saturated solution where an equilibrium exists between the undissolved solute and the solute particles in solution.
12. (C) A 10.% solution contains 10. g of solute/100. g of solution because the percentage is by mass/100 g of solution.
13. (D) The molar mass is equal to the total of Al = 27 + 3Cl = 106.5 for a total of 133.5 g/mol.
14. (C) 98 g H2SO4 = 1 mol H2SO4
So, if there are 500 mL (which is 0.5 L) then 1 mol in 0.5 L would be equivalent to 2 mol in 1 L or a 2 molar solution.
15. (B) 684 g of sucrose = = 2 mol of sucrose
So = 1 molal solution (1 m)
The freezing point is lowered to –1.86°C
16. (A) The graph of the solubility of a substance that has no change in its solubility as the temperature is increased would be a straight line, as shown in A.
17. (C) The graph that shows the amount of solute in solution as it is added up through its saturation point is C. The point at which saturation occurs is where the line flattens out to a horizontal straight line.
18. (D) Graph D shows the ice getting warmer until it starts to melt. This is where the temperature stays the same until it is all melted. The heat being added is being used for the change in state from a solid to a liquid and the line stays horizontal, indicating no change in temperature during this time. The line then starts to ascend as liquid water is absorbing the heat and increasing in temperature.
19. (C) The molar mass of Ca(OH)2 is Ca = 40, 2 O = 32, and 2 H = 2, which, added together equals 74 g/mol. If this is dissolved to make a 2 L solution, there is only .5 mol in 1 L. So the molarity (M), the moles of solute in 1 L = .5 M
20. (B) The molar mass of C12 H22 O11 is 12 C = 144, 22 H = 22, and 11 O = 176, which added together equals 342 g/mol. Because 684 g is given, then 684 g divided by 342 g/mol = 2 mol of sucrose. If this is dissolved in 2,000 g of water, it gives 1 mol /1,000 g of water or a 1 molal solution.