MC Unit 7 Practice Exam Questions. Multiple ... What is the molarity of a solution
that contains 54 g KCl in 2.50 L ... Solutions with water as the solvent are called.
MC Unit 7 Practice Exam Questions Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. 9. The rate at which a solid solute can be dissolved in a liquid solvent can be increased by a. lowering the temperature of the solvent. b. grinding the solute into small pieces. c. increasing the air pressure on the liquid. d. lowering the temperature of the solute.
1. Which of the following does NOT increase the rate of dissolving a solid in water? a. increasing the kinetic energy b. increasing surface area c. decreasing surface area d. using smaller pieces of solid 2. Which of the following will dissolve most slowly? a. large salt crystals in unstirred water b. large salt crystals in stirred water c. small salt crystals in unstirred water d. small salt crystals in stirred water 3. Raising solvent temperature causes solvent-solute collisions to become a. less frequent and more energetic. b. more frequent and more energetic. c. less frequent and less energetic. d. more frequent and less energetic. 4. Which of the following expresses concentration? a. parts per million b. molarity c. percent concentration by mass d. all of the above 5. What is the molarity of a solution that contains 54 g KCl in 2.50 L solution? a. 5.3 M b. 0.29 M c. 22 M d. 0.47 M
Figure 15-1
6. Solutions with water as the solvent are called a. liquids. b. miscible solutions. c. aqueous solutions. d. electrolytic.
10. How would you describe the level of saturation of a solution of KCl at 90°C in Figure 15-1 if there are 55 grams of NaCl in 100 grams of water? a. It is supersaturated. b. It is dilute. c. It is unsaturated. d. It is saturated.
7. Molarity is expressed as a. moles of solvent/liters of solute. b. moles of solute/liters of solvent. c. moles of solute/liters of solution. d. moles of solute/moles of solvent.
11. Using Figure 15-1, determine how many grams of KClO3 needs to dissolve in 200 grams of water in order to have a saturated solution at 50°C. a. 140 grams b. 70 grams c. 18 grams d. 36 grams
8. Which of the following's solubility is most affected by pressure? a. ionic solids b. supersaturated solutions c. gases d. alloys
12. Which substance in Figure 15-1 is the most soluble in water at 0 degrees C? a. NaNO3 b. KClO3 c. NH3 d. None are soluble at that temperature.
Free Response/Short Answer 13. A solution contains 85.0 g of NaNO3, and has a volume of 750. mL. Find the molarity of the solution.
14. How many grams of Na2SO4 are needed to prepare 750. mL of a 0.375 M solution?
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15. Distinguish among the terms saturated, unsaturated, dilute, and concentrated. 16. Explain why crushed salt (NaCl) dissolves faster in water than large lumps of rock salt. 17. What is the molarity of a solution prepared by dissolving 5.68 grams of NaOH in enough water to make 400. mL of solution? 18. How many grams of HCl are required to prepare 250 milliliters of a 0.158 M solution?
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ID: A
MC Unit 7 Practice Exam Questions Answer Section MULTIPLE CHOICE 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.
85.0 g NaNO 3 1 mol NaNO 3 1000 mL × × = 1.33 M NaNO 3 750. mL 85.0 g NaNO 3 1L PTS: 1 14. ANS: 40.0 g Na2SO4
750. mL ×
0.375 mol Na 2 SO 4 142.05 g Na 2 SO 4 1L × × = 40.0 g Na 2 SO 4 1000 mL 1L 1 mol Na 2 SO 4
PTS: 1 15. ANS: Saturated and unsaturated are related to the amount of solute dissolved in a solvent compared to how much can be dissolved. Thus, a saturated solution may have a large or small amount of solute dissolved, depending on the solubility of the substance. Dilute and concentrated refer to the actual amount of solute dissolved in a solvent. A concentrated solution always has a large amount of solute, and a dilute solution always has a small amount of solute. PTS: 1 16. ANS: Crushed salt will have a larger surface area than a large lump of salt, thus enabling the water molecules to come in contact more quickly with a large number of Na+ and Cl- ions, causing solution. PTS:
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ID: A
17. ANS:
1 mol = 0.142 mole NaOH 40 g 0.142 mol NaOH = 0.355 M 0.4 L
5.68 g ×
PTS: 1 18. ANS: 158 mol/L × 0.250 L × 6.5 g/mol = 1.44 g PTS:
