Ch 13 Notes Solutions Concentration

Concentration: the concentration of a solution is a measure of the amount of solute in a given amount of solvent or solution. I. Molarity: the number of moles of solute in one liter of solution.

M = amount of solute (moles) OR grams of solute/molar mass of solute volume of solution ( liters) L of solution

A bottle labeled 6M HCl is pronounced 6 molar HCl and it was prepared by mixing 6 moles of HCl with enough water to make 1 liter of solution. Molarity is always moles/liter. Ex: 6M is 6 moles/ liter

Example Molarity Problems: 1. Calculate the molarity, M, of a solution prepared by dissolving 11.85 g of potassium permanganate in enough water to make 750. mL of solution.

2. Calculate the mass of NaCl needed to prepare 175 ml of 0.500 M saline solution.

3. Calculate the volume (in mL) needed to prepare a 2.48 M sodium hydroxide solution containing 31.52 g of the dissolved solid.

4. How many grams of calcium chloride must be dissolved in water to make 350. mL of a l.75 M solution?

5. What is the molar mass of 55.0 grams of a solute that has been dissolved in enough solvent to make 500. mL of a l.5 M solution?

II. Dilutions – you need to make 800.0 mL of a 0.25 M solution of HCl. The only available HCl is concentrated (12 M). How would you do this? Being able to prepare dilutions is a very common application of chemistry. Our department buys concentrated acids, but normally uses more dilute solutions of these acids in our labs. Therefore, it is important to know how to correctly dilute.

M1V1 = M2V2 where M1 = concentrated solution V1 = volume of concentrated solution M2 = dilute solution V2 = volume of dilute solution

In the above problem, M1 = 12.0 M M2 = 0.25 M Vl = ? V2 = 800.0 ml therefore

(12.0 M) (V1) = (0.25 M)(800 mL) and V1 = 16.67 ml.

This 16.67 mL is the amount of concentrated acid we will take out, but we still have to make 800 mL of 0.25 M. The other 783.3 mL of solution must be water (800 – 16.67). Therefore we would place 16.67 mL of HCl into 783.3 mL of water and we would have our diluted solution.

Practice Dilutions Problems: 6. How would you prepare 485 mL of 0.39 M solution of NaCl when a l.0 M solution of NaCl is all you can find?

7. If 300.0 mL of a 2.5 M solution of nitric acid is added to 500.0 mL of water, what is the molarity of the dilute solution?

8. Prepare 500. mL of a dilute solution (0.50 M ) of nitric acid from the l5.0 M stock solution. You will need 600. mL of the dilute solution.

9. How would you prepare 500. mL of a 0.250 M solution of NaCl from a 3.00 M stock solution?

10. Tell how you would prepare enough 0.75 M NaCl solution so that 78 students working in groups of 2 will have 12 mL of solution for a lab. The stock solution is 3.5 M NaCl.

III. Percent Solutions (2 types): Percent mass or %(m/m)– used when a solid solute is dissolved in liquid, usually water.

% (m/m) = grams of solute grams of solution

11. Prepare a 10.00 % NaCl solution using 50.0 g water and solid salt:

12. How many grams of water must be added to 25.0 g salt in order to have a 2.00 % (by mass) salt solution?

13. Prepare 600.0 g of a 3.00 % saline solution (NaCl solution).

Percent volume or %(v/v) – used when a liquid solute is mixed with a liquid solvent. The units are mL or L, but are worked the same. % (v/v) = mL solute mL solution

14. Prepare a 20.00 % alcohol solution using 400.0 mL of water:

15. What is the percent (v/v)of ethanol in the final solution when 90.0 mL of it are diluted to a volume of 300. mL with water?