SOLUTIONS WORKSHEET

Introduction

Molarity and molality of solutions are explained in most standard chemistry texts, and taught in most standard chemistry courses. In biology, however, solutions are more often written as a percentage rather than as a molarity or molality. (This keeps the biologist from having to add up molecular masses and doing any complex math.) Thus, while molarity and molality are often covered in a chemistry class, solutions as a percentage is not normally taught.

The concept is very straightforward. A 1% solution means there is 1 gram of solute dissolved in 99 grams of solvent. Thus, the solute makes up 1% of the total final mass of the solution.

As an example, if it is desired to prepare 50.0 grams of a 2.0% NaCl solution the following calculations would be made.

Total mass of solute and solvent = 50.0 grams

Mass of solute (NaCl) = 50.0 g X 2% = 1.0 g

Mass of solvent (water) = 50.0 g - 1.0 grams = 49.0 g

So the solution would be prepared by dissolving 1.0 gram of NaCl in 49.0 grams of water.

Student Objectives

* Students will correctly determine the number of grams of solute and solvent given its percent of solution and the total final mass of the solution

* Students will review how to make solutions based on molarity and molality

* Students will review the concept of density

* Students will practice conversions from cm³ to dm³.

(1000 cm³ = 1000 ml = 1 dm³ = 1 liter)

Materials

Paper, pencil and calculator

Procedure

1. Answer the following, show all work.

Molarity is the number of of solute divided by the number of of solvent.

The abbreviation for molarity is .

Molality is the number of of solute divided by the number of

of solvent.

The abbreviation for molality is .

% solution means the solvent is of the total

of the final solution.

Assume that all solutions are being dissolved in water.

Assume that the density of water is 1.0 g/cm3 and that the final density of all solutions is 1.0 g/cm³. (Thus, 1 cm³ of solution equals 1 gram of solution.)

1. How would you make 5.00 X 10² cm³ of a 4.0 M NaCl solution?

2. How would you make 1.00 X 10³ cm³ of a 5.5 m KBr solution?

3. How would you make 40.0 grams of a 5.0% KOH solution?

4. How would you make 40.0 cm3 of a 2.0 % Mg(OH)₂ solution?

5. How would you make 50.0 cm³ of a 0.025 M NaF solution?

6. How would you make 100.0 cm³ of a 0.50 m KCl solution?

7. How would you make 400.0 grams of a 7.0% Sr(OH)₂ solution?

8. How would you make 400.0 cm³ of a 4.0 % KNO₃ solution?

9. How would you make 4.0 grams of a 1.0% Na₂SO₄ solution?

10. How would you make 70.0 dm³ of a 25 % Ba(NO₃)₂ solution?

11. How would you make 17.0 grams of a 2.0% sucrose solution?

12. How would you make 0.250 dm³ of a 2.50 % Ca(NO₃)₂ solution?

13. How would you make 0.050 dm³ of a 1.0 % agarose?

14. How would you make 0.050 dm³ of a 2.0 % agarose?

15. How would you make 0.050 dm³ of a 3.0 % agarose?

16. How would you make 0.050 dm³ of a 4.0 % agarose?

17. How would you make 50.0 cm³ of a 1.0 % agarose?

18. How would you make 50.0 cm³ of a 2.0 % agarose?

19. How would you make 50.0 cm³ of a 3.0 % agarose

20. How would you make 50.0 cm³ of a 4.0 % agarose?

The formula for an agarose monomer is C₁₂H₁₈O₉

21. What is the molecular mass of agarose?

22. Determine the molarity of a 1% agarose solution. (Hint: assume you are making 100 grams of total solution.)

23. Determine the molarity of a 2% agarose solution. (Hint: assume you are making 100 grams of total solution.)

24. Determine the molarity of a 3% agarose solution. (Hint: assume you are making 100 grams of total solution.)

25. Determine the molarity of a 4% agarose solution. (Hint: assume you are making 100 grams of total solution.)