MU #1 The concentration of a solution may be expressed as molarity (M), percent by volume, percent by mass, or parts per million (ppm).

percent by mass

See table T for percent composition.

% composition by mass = (mass of part / mass of whole) x 100

Example

Consider 10 g of sucrose dissolved in 40 g of water.

The % composition by mass of sucrose = (10g / 50 g) x 100 = 20% sucrose

Therefore, the mixture is composed of 20% of sucrose and 80% is water.

percent by volume

See table T for percent composition.

Replace mass by volume in the formula

% composition by volume = (volume of part / volume of whole) x 100

Example

Consider 1L of O₂(g) dissolved in 9L of water.

The % composition by volume of O₂(g) = (1L / 9L+1L) x 100 = 10% O₂(g)

Therefore, the mixture is composed of 10% of O₂(g) and 90% of water.

parts per million (ppm)

See table T for parts per million.

parts per million = (grams of solute / grams of solution) x 1,000,000

Example

Consider 0.0010g of CO₂(g) dissolved in 1,000g of water.

The part per million of CO₂(g) = (0.0010g / 1,000g) x 1,000,000 = 1 ppm of CO₂(g)

molarity (M)

See table T molarity (M)

molarity (M) = moles of solute / liters of solution

Example

Consider 2 moles of KCl(s) dissolved in 100ml of water. Remember to convert ml into L (1,000ml = 1L)

The molarity of the solution = 2 mol / 0.1L = 20 mol/L = 20 M KCl(aq)

Therefore, there are 20 moles of KCl(s) in every liter of water.

To convert moles into grams use the mole calculation formula in table T.

number of moles = given mass (g) / gram-formula mass

Example

Consider converting 2 moles of KCl(s) into grams.

2 mol of KCl(s) = given mass (g) of KCl(s) / gram-formula mass of KCl

gram-formula mass = atomic mass (rounded to a whole number) in g/mol of all the elements in the formula

See periodic table of elements

gram-formula mass of KCl(s) = mass of 1 K + mass of 1 Cl = 39g/mol + 35g/mol = 74g/mol

Therefore, 2 mol of KCl(s) = given mass (g) of KCl(s) / 74g/mol

Then, given mass (g) of KCl(s) = (2 mol) (74g/mol) = 148g of KCl(s)

Solutions of a certain concentration or molarity are prepared the following way

Step 1 Calculate the number of moles of solute using the formula for molarity  in table T.

            molarity = moles of solute / liters of solution

Step 2  Convert the moles of solute into grams using the formula for mole calculation in table T.

            number of moles = given mass (g) / gram-formula mass

Step 3 Add the grams of solute in the liters of solution.

Example

Consider the making of 200ml of 2M NaCl(aq).

Step 1 Calculate the number of moles of NaCl(s)

            2 mol/L = moles of  NaCl(s) / 0.2L

            Therefore, moles of NaCl(s) = (2 mol/L) (0.2L) = 0.4 mol of NaCl(s) 

Step 2  Convert the moles of NaCl(s) into grams of NaCl(s).

            0.4 mol of NaCl(s) = given mass (g) NaCl(s) / gram-formula mass

            gram-formula mass of NaCl = mass of 1 Na + mass of 1 Cl = 23g/mol + 35g/mol = 58g/mol

            Therefore, 0.4 mol of NaCl(s) = given mass (g) of NaCl(s) / 58g/mol

            Then, given mass (g) of NaCl(s) = (0.4 mol) (58g/mol) = 23g of NaCl(s)

Step 3 Add 23g of NaCl(s) in 0.2 L of H₂O(l)

Problems

1. What is the molarity of a solution of KOH if 500 ml of the solution contains 5.6 grams of KOH?

2. How many moles of NaOH are contained in 200 mL of 0.l M solution of NaOH?

3. How many moles of KOH are contained in 500 mL of 0.5 M solution of KaOH?

TEST YOUR UNDERSTANDING

1. What is the concentration of a solution of 10 moles of copper (II) nitrate in 5.0 liters of solution?

            1. 0.50 M         2. 2.0 M           3. 5.0 M           4. 10. M

2. A 20.-milliliter sample of 0.60 M HCl is diluted with water to a volume of 40. milliliters. What is the new concentration of the solution?

            1. 0.15 M         2. 0.60 M         3. 0.30 M         4. 1.2 M

3. A student obtained the following data to determine the percent by mass of water in a hydrate. What is the approximate percent by mass of the water in the hydrated salt?

            1. 2.5%            2. 12%             3. 88%             4. 98%

6/02

What is the molarity of a solution that contains 0.50 mole of NaOH in 0.50 liter of solution? (1) 1.0 M  (2) 2.0 M  (3) 0.25 M  (4) 0.50 M

8/02

1 A hydrated salt is a solid that includes water molecules within its crystal structure. A student heated a 9.10-gram sample of a hydrated salt to a constant mass of 5.41 grams. What percent by mass of water did the salt contain?

            (1) 3.69%      (2) 16.8%       (3) 40.5%         (4) 59.5%

2  How many moles of solute are contained in 200 milliliters of a 1 M solution?  (1) 1       (2) 0.2     (3) 0.8     (4) 200

3  What is the total number of grams of NaI(s) needed to make 1.0 liter of a 0.010 M solution?  (1) 0.015    (2) 0.15     (3) 1.5   (4) 15

1/03

What is the molarity of a solution of NaOH if 2 liters of the solution contains 4 moles of NaOH?  (1) 0.5 M   (2) 2 M     (3) 8 M   (4) 80 M

1/04

41 What is the molarity of a solution containing 20 grams of NaOH in 500 milliliters of solution? (1) 1 M (2) 2 M (3) 0.04 M (4) 0.5 M

8/04

53 In the space provided in your answer booklet, show a correct numerical setup for determining how many liters of a 1.2 M solution can be prepared with 0.50 mole of C₆H₁₂O₆.  

61 What is the gram-formula mass of (NH₄)₂CO₃? Use atomic masses rounded to the nearest whole number.

1/05

78 What is the gram-formula mass of ethylene glycol, C₂H₄(OH)₂?

79 Show a correct numerical setup for calculating the total number of moles of ethylene glycol needed to prepare 2.50 liters of a 10.0 M solution.

6/05

42 What is the concentration of a solution, in parts per million, if 0.02 gram of Na3PO4 is dissolved in 1000 grams of water?  

		(1) 20 ppm   (2) 2 ppm (3) 0.2 ppm (4) 0.02 ppm 

1/06

6/06

8/06

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