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Chapter 7 Solutions

7.5 Molarity and Dilution

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Molarity (M)

Molarity (M) is

• a concentration term for solutions.

• gives the moles of solute in 1 L solution.

• moles of soluteliter of solution

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Preparing a 1.0 Molar Solution

A 1.00 M NaCl solution is prepared• by weighing out 58.5 g NaCl (1.00 mole) and• adding water to make 1.00 liter of solution.

Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings

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What is the molarity of 0.500 L NaOH solution if itcontains 6.00 g NaOH?

STEP 1 Given 6.00 g NaOH in 0.500 L solution Need molarity (mole/L)

STEP 2 Plan g NaOH mole NaOH molarity

Calculation of Molarity

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Calculation of Molarity (cont.)

STEP 3 Conversion factors 1 mole NaOH = 40.0 g1 mole NaOH and 40.0 g NaOH40.0 g NaOH 1 mole NaOH

STEP 4 Calculate molarity.6.00 g NaOH x 1 mole NaOH = 0.150 mole

40.0 g NaOH 0.150 mole = 0.300 mole = 0.300 M NaOH

0.500 L 1 L

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What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO3?

1) 0.557 M 2) 1.44 M3) 1.71 M

Learning Check

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3) 1.71 M 46.8 g NaHCO3 x 1 mole NaHCO3 = 0.557 mole NaHCO3

84.0 g NaHCO3

0.557 mole NaHCO3 = 1.71 M NaHCO3

0.325 L

Solution

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What is the molarity of 225 mL of a KNO3 solution containing 34.8 g KNO3?

1) 0.344 M

2) 1.53 M

3) 15.5 M

Learning Check

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2) 1.53 M34.8 g KNO3 x 1 mole KNO3 = 0.344 mole KNO3

101.1 g KNO3

M = mole = 0.344 mole KNO3 = 1.53 M L 0.225 LIn one setup:

34.8 g KNO3 x 1 mole KNO3 x 1 = 1.53 M 101.1 g KNO3 0.225 L

Solution

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Molarity Conversion Factors

The units of molarity are used as conversion factors in calculations with solutions.

TABLE 7.8

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Calculations Using Molarity

How many grams of KCl are needed to prepare 125 mLof a 0.720 M KCl solution?

STEP 1 Given 125 mL (0.125 L) of 0.720 M KCl Need Grams of KCl

STEP 2 Plan L KCl moles KCl g KCl

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Calculations Using Molarity

STEP 3 Conversion factors 1 mole KCl = 74.6 g1 mole KCl and 74.6 g KCl 74.6 g KCl 1 mole KCl

1 L KCl = 0.720 mole KCl 1 L and 0.720 mole KCl0.720 mole KCl 1 L

STEP 4 Calculate grams.0.125 L x 0.720 mole KCl x 74.6 g KCl = 6.71 g KCl 1 L 1 mole KCl

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How many grams of AlCl3 are needed to prepare 125 mL of a 0.150 M solution?

1) 20.0 g AlCl3

2) 16.7g AlCl3

3) 2.50 g AlCl3

Learning Check

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Solution

3) 2.50 g AlCl3

0.125 L x 0.150 mole x 133.5 g = 2.50 g AlCl3 1 L 1 mole

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How many milliliters of 2.00 M HNO3 contain 24.0 g HNO3?

1) 12.0 mL

2) 83.3 mL

3) 190. mL

Learning Check

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24.0 g HNO3 x 1 mole HNO3 x 1000 mL =

63.0 g HNO3 2.00 mole HNO3

Molarity factor inverted

= 190. mL HNO3

Solution

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Dilution

In a dilution• water is added.• volume increases.• concentration decreases.

MixThe amount of solute in the concentrated solution = amount of solute in the diluted solution

3 containers of H2O

1 container orange juice concentrate

+

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Comparing Initial and Diluted Solutions

In the initial and diluted solution,• the moles of solute are the same.• the concentrations and volumes are related by

the following equations:For percent concentration:C1V1 = C2V2

initial diluted

For molarity:M1V1 = M2V2

initial diluted

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Dilution Calculations with Percent

What volume of a 2.00% (m/v) HCl solution can be prepared by diluting 25.0 mL of 14.0% (m/v) HCl solution?Prepare a table:

C1= 14.0% (m/v) V1 = 25.0 mL

C2= 2.00% (m/v) V2 = ?

Solve dilution equation for unknown and enter values:

C1V1 = C2V2

V2 = V1C1 = (25.0 mL)(14.0%) = 175 mL

C2 2.00%

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Learning Check

What is the percent (% m/v) of a solution prepared by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?

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Solution

What is the percent (%m/v) of a solution prepared by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?Prepare a table:

C1= 9.00 %(m/v) V1 = 10.0 mL

C2= ? V2 = 60.0 mL

Solve dilution equation for unknown and enter values:

C1V1 = C2V2

C2 = C1 V1 = (10.0 mL)(9.00%) = 1.50% (m/v)

V2 60.0 mL

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Dilution Calculations with Molarity

What is the molarity (M) of a solution prepared by diluting 0.180L of 0.600 M HNO3 to 0.540 L?

Prepare a table:M1= 0.600 MV1 = 0.180 L

M2= ? V2 = 0.540 L

Solve dilution equation for unknown and enter values:

M1V1 = M2V2

M2 = M1V1 = (0.600 M)(0.180 L) = 0.200 M

V2 0.540 L

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Learning Check

What is the final volume (mL) of 15.0 mL of a 1.80 MKOH diluted to give a 0.300 M solution?

1) 27.0 mL2) 60.0 mL 3) 90.0 mL

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Solution

What is the final volume (mL) of 15.0 mL of a 1.80 MKOH diluted to give a 0.300 M solution?Prepare a table:

M1= 1.80 M V1 = 15.0 mL

M2= 0.300 M V2 = ?

Solve dilution equation for V2 and enter values:

M1V1 = M2V2

V2 = M1V1 = (1.80 M)(15.0 mL) = 90.0 mL

M2 0.300 M