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13.E: Homework Chapter 13 Answers

  • Page ID
    202224
  • Solutions, Solvents, and Solutes:

    1. a.) Solution: a homogeneous mixture composed of two or more substances

    b.) Solvent: the major component in a solution

    c.) Solute: the minor component in a solution

    3. The two substances will not mix together. Imagine mixing oil and water together, the oil will float to the top of the water, and not mix together. This would not be a solution, this would be a heterogeneous mixture.

    5. Solvent: ethyl alcohol; Solute: water

    7. Solvent: water; Solute: sodium hydroxide

    9. c.) CH3CH2CH2CH3      *remember: “like dissolves like,” butane is nonpolar like grease*

    11. 25.245g KNO3  

    13. 19.912 g CO2

    15. 2.0812 mol Al; Aluminum is the solute.

    17.  The solution would be unsaturated, because at 25 degrees Celsius, the point would fall below the solubility curve, indicating that it is unsaturated.

    Mass Percent Calculations:

    19. = 33.7 %

    21. a.) 4.52%

    b.) 10.9%

    c.) 9.09%

    23. 107%

    25. 1.8%

    27. 29.7%

    29. 6.08%

    Molarity Calculations:

    31. a.) 0778 M

    b.) 0.108 M

    c.) 0.0521 M

    d.) 0.493 M

     Text Box: **Remember: you must first convert what is given to you to moles and liters to find the molarity**

     

     

    33. 0.255 g KCl

    35. 0.752 M

    37. 0.0344 M

    39. 7.68 M

    Dilution Calculations:

    41. M2 = 7.05 M

    43. V1 = 12.5 mL

    45. You would need to take 3.00mL of the 10.0 M stock solution and dilute it to 15.0 mL.

    47. a.) V2 = 18.0 mL

    b.) V2 = 14.0 mL

    c.) V2 =52.5 mL

    d.) V2 = 40.0 mL

    49. V2 = 18.545 L

    51. V2 = 125 mL

    53. a.) M1 = 5.00 M

    b.) M1 = 6.00 M

    c.) M1 = 7.50 M

    d.) M1 = 5.7692 M

    55. M1 = 3.50 M

    57. V2 = 24.0 mL

    Cumulative/Challenge Problems:

    59. Solvent: soap; Solute: water

    61. M2 = 1.67 M