4.1.3 Concentrations of Solutions

Question 1

Concentrations of Solutions

Answer 1

• The concentration of a solution is measured as the amount of solute per volume.
• Knowing the concentration of a solution allows one to calculate the concentrations of solutions made by diluting the original solution.
• Several different concentration units are used by chemists.

Question 2

molarity

Answer 2

• The concentration of a solution is measured as the amount of solute per volume.
• Molarity is the concentration unit most frequently used by chemist. One molar is one mole of solute per liter of solution.
• A solution with a known molarity can be prepared by dissolving a known mass of solute in enough water to fill a 200 mL volumetric flask. The molarity of this solution can be calculated by converting the mass of solute to moles of solute (using the molar mass of the solute) and dividing by the volume in liters.
• The moles of a given ion in a given quantity of solution can be calculated knowing the stoichiometry of the solute. For example, every mole of sodium sulfate yields two moles of sodium ion (Na+). Therefore, 1.86 moles of sodium ion are present in one liter of a 0.930 M solution of sodium sulfate

Question 3

dilution

Answer 3

• Dilutioninvolves making a less concentrated solution by adding solvent to a given amount of a more concentrated solution. Knowing the concentration of a solution allows one to calculate the concentrations of solutions made by diluting the original solution.
• Problem: How much of a 0.930 M solution is needed to make 250. mL of a 0.10 M Na2SO4 solution?
• This problem can be solved by noting that the number of moles of solute present before diluting is equal to the number of moles present after the solution is diluted. The number of moles is equal to the molarity (M) multiplied by the volume (V). Therefore, M1V1 = M2V2. Solving for V1 yields the volume of the original solution required to make the desired solution.

Question 4

You have 250 mL of a 0.80 M solution of silver nitrate AgNO3 (169.87 g / mol) dissolved in water. How many grams of AgNO3 does the solution contain?

Answer 4

34 g

Question 5

Which of the following describes the best way to prepare 500. mL of a 0.20 M solution of benzene (C6H6 ) (78.11 g / mol) dissolved in carbon tetrachloride (CCl4 ) (153.82 g / mol)?

Answer 5

Add 7.8 g of C6H6 to a volumetric flask. Then fill the flask to the 500 mL line with CCl4.

Question 6

If 30. g of potassium iodide KI (166.00 g / mol) is dissolved in 250 g of water, what is the molality of this solution?

Answer 6

0.72 molal

Question 7

Under what circumstances would it be most appropriate to express the concentration of a solute in parts per million?

Answer 7

When the solution is very dilute.

Question 8

You mix 3.4 g of potassium chromate K2CrO4 (194.19 g / mol) with water. If the resulting solution has a volume of 150 mL, what is its molar concentration?

Answer 8

0.12 M

Question 9

Suppose 50.0 g of alcohol are dissolved in 230. g of water. What is the percent of alcohol by mass of the solution?

Answer 9

17.9%

Question 10

Suppose 26 g of calcium chloride CaCl2 (110.98 g / mol) are dissolved in water to make 200 mL of solution. How many times greater is the molarity of chloride ions (Cl − ) than the molarity of calcium ions (Ca2+ ) in the solution?

Answer 10

2.0

Question 11

You have 500 mL of a 0.5 M solution of ammonia NH3 (17.03 g / mol) dissolved in water. If you pour a 20 mL sample into a beaker, what will the molar concentration of the sample be?

Answer 11

equal to 0.5 M

Question 12

You have 50. mL of a 1.7 M solution of acetic acid dissolved in water. Approximately how many milliliters of water must be added to dilute it to a 0.20 M solution? (assume volumes are additive)

Answer 12

380 mL

Question 13

You have a 12 M solution of hydrochloric acid HCl (36.5 g / mol). How many milliliters of the original solution do you need in order to make 500 mL of a 0.50 M solution of HCl?

Answer 13

21 mL