Solutions to Chapter 5 Exercises

5.1 Solutions

1. (a) 12.01 amu; (b) 12.01 amu; (c) 144.12 amu; (d) 60.05 amu

3. (a) 123.896 amu; (b) 18.015 amu; (c) 164.086 amu; (d) 60.052 amu; (e) 342.297 amu

5. (a) 56.107 amu;
(b) 54.091 amu;
(c) 199.9976 amu;
(d) 97.9950 amu

7. Use the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of compound by the molar mass of the compound expressed in grams.

9. Formic acid. Its formula has twice as many oxygen atoms as the other two compounds (one each). Therefore, 0.60 mol of formic acid would be equivalent to 1.20 mol of a compound containing a single oxygen atom.

11. The two masses have the same numerical value, but the units are different: The molecular mass is the mass of 1 molecule while the molar mass is the mass of 6.022 × 10²³ molecules.

13.(a) 256.528 g/mol; (b) 72.150 g mol⁻¹; (c) 378.103 g mol⁻¹; (d) 58.080 g mol⁻¹; (e) 180.158 g mol⁻¹

15. (a) 197.382 g mol⁻¹; (b) 257.163 g mol⁻¹; (c) 194.193 g mol⁻¹; (d) 60.056 g mol⁻¹; (e) 306.464 g mol⁻¹

17. (a) 0.819 g;
(b) 307 g;
(c) 0.23 g;
(d) 1.235 × 10⁶ g (1235 kg);
(e) 765 g

19. (a) 99.41 g;
(b) 2.27 g;
(c) 3.5 g;
(d) 222 kg;
(e) 160.1 g

21. (a) 9.60 g; (b) 19.2 g; (c) 28.8 g

23. zirconium: 2.038 × 10²³ atoms; 30.87 g; silicon: 2.038 × 10²³ atoms; 9.504 g; oxygen: 8.151 × 10²³ atoms; 21.66 g

25. AlPO₄: 1.000 mol
Al₂Cl₆: 1.994 mol
Al₂S₃: 3.00 mol

27. 3.113 × 10²⁵ C atoms

29. 0.865 servings, or about 1 serving.

31. 20.0 g H₂O represents the least number of molecules since it has the least number of moles.

5.2 Solutions

1. We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.

3. (a) 0.679 M;
(b) 1.00 M;
(c) 0.06998 M;
(d) 1.75 M;
(e) 0.070 M;
(f) 6.6 M

5. (a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass; (b) 27 g

7. (a) 37.0 mol H₂SO₄;
3.63 × 10³ g H₂SO₄;
(b) 3.8 × 10⁻⁶ mol NaCN;
1.9 × 10⁻⁴ g NaCN;
(c) 73.2 mol H₂CO;
2.20 kg H₂CO;
(d) 5.9 × 10⁻⁷ mol FeSO₄;
8.9 × 10⁻⁵ g FeSO₄

9. (a) Determine the molar mass of KMnO4; determine the number of moles of KMnO₄ in the solution; from the number of moles and the volume of solution, determine the molarity; (b) 1.15 × 10⁻³ M

11. (a) 5.04 × 10⁻³ M;
(b) 0.499 M;
(c) 9.92 M;
(d) 1.1 × 10⁻³ M

13. 0.025 M

15. 0.5000 L

17. 1.9 mL

19. (a) 0.125 M;
(b) 0.04888 M;
(c) 0.206 M;
(e) 0.0056 M

21. 11.9 M

23. 1.6 L

5.3 Solutions

**1. (a) The dilution equation can be used, appropriately modified to accommodate mass-based concentration units:
%mass₁ × mass₁ = %mass₂ × mass₂
This equation can be rearranged to isolate mass1 and the given quantities substituted into this equation.
(b) 58.8 g

3. 114 g

5. 1.75 × 10⁻³ M

7. 95 mg/dL

9. 2.38 × 10⁻⁴ mol

11. 0.29 mol