Chapter 8 Solutions

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# Chapter 8 Solutions - PowerPoint PPT Presentation

Chapter 8 Solutions. 8.5 Molarity and Dilution. Molarity (M). Molarity (M) is a concentration term for solutions gives the moles of solute in 1 L solution molarity (M) = moles of solute liter of solution. Preparing a 1.0 Molar NaCl Solution.

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Chapter 8 Solutions

8.5 Molarity and Dilution

Molarity (M)

Molarity (M)

• is a concentration term for solutions
• gives the moles of solute in 1 L solution

molarity (M) = moles of solute

liter of solution

Preparing a 1.0 Molar NaCl Solution

A 1.0 M NaCl solution is prepared

• by weighing out 58.5 g of NaCl (1.00 mole) and
• adding water to make 1.0 liter of solution
Example of Calculating Molarity

What is the molarity of 0.500 L NaOH solution if it

contains 6.00 g of NaOH?

STEP 1 Given: 6.00 g of NaOH in 0.500 L solution

Need: molarity (mole/L) of NaOH solution

STEP 2 Plan:

g of NaOH moles of NaOH molarity

Example of Calculating Molarity (continued)

STEP 3Write equalitites and conversion factors:

1 mole of NaOH = 40.0 g of NaOH

1 mole NaOH and 40.0 g NaOH

40.0 g NaOH 1 mole NaOH

STEP 4 Set up problem to calculate molarity:

6.00 g NaOH x 1 mole NaOH = 0.150 mole of NaOH

40.0 g NaOH

0.150 mole NaOH = 0.300 mole NaOH

0.500 L NaOH solution 1 L NaOH solution

= 0.300 M NaOH

Learning Check

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

1) 0.557 M NaHCO3

2) 1.44 M NaHCO3

3) 1.71 M NaHCO3

Solution

3) 1.71 M

STEP 1 Given: 46.8 g of NaHCO3

325 mL (0.325 L) NaHCO3 solution

Need: molarity (mole/L) of NaHCO3 solution

STEP 2 Plan:

g of NaHCO3 moles of NaHCO3 molarity

STEP 3Write equalities and conversion factors:

1 mole of NaHCO3 = 84.0 g of NaHCO3

1 mole NaHCO3 and 84.0 g NaHCO3

84.0 g NaHCO3 1 mole NaHCO3

Solution (continued)

STEP 4 Setup problem to calculate moles and molarity of NaHCO3:

46.8 g NaHCO3 x 1 mole NaHCO3

84.0 g NaHCO3

= 0.557 mole of NaHCO3

0.557 mole NaHCO3 = 1.71 mole NaHCO3

0.325 L NaHCO3 solution 1 L NaHCO3 solution

= 1.71 M NaHCO3

Learning Check

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

1) 0.344 M

2) 1.53 M

3) 15.5 M

Solution

2) 1.53 M KNO3

STEP 1 Given: 34.8 g of KNO3

225 mL (0.225 L) KNO3 solution

Need: molarity (mole/L) of KNO3 solution

STEP 2 Plan:

g of KNO3 moles of KNO3 molarity

STEP 3Write equalities and conversion factors:

1 mole of KNO3 = 101.1 g of KNO3

1 mole KNO3 and 101.1 g KNO3

101.1 g KNO3 1 mole KNO3

Solution (continued)

STEP 4 Set up problem to calculate moles and molarity of KNO3:

34.8 g KNO3 x 1 mole KNO3

101.1 g KNO3

= 0.344 mole of KNO3

0.344 mole KNO3 = 1.53 mole KNO3

0.225 L KNO3 solution 1 L KNO3 solution

= 1.53 M KNO3

In one setup:

34.8 g KNO3 x 1 mole KNO3x 1 = 1.53 M

101.1 g KNO3 0.225 L

Molarity Conversion Factors

The units of molarity are used to write conversion factors for calculations with solutions.

Example of Calculations Using Molarity

How many grams of KCl are needed to prepare 125 mL

of 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 of KCl moles of KCl g of KCl

Example of Calculations Using Molarity (continued)

STEP 3Write equalities and conversion factors:

1 mole of KCl = 74.6 g of KCl

1 mole KCl and 74.6 g KCl

74.6 g KCl 1 mole KCl

1 L of KCl = 0.720 mole of KCl

1 L and 0.720 mole KCl

0.720 mole KCl 1 L

STEP 4 Set up problem to cancel mole KCl:

0.125 L x 0.720 mole KCl x 74.6 g KCl = 6.71 g of KCl

1 L 1 mole KCl

Learning Check

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

1) 20.0 g of AlCl3

2) 16.7 g of AlCl3

3) 2.50 g of AlCl3

Solution

3) 2.50 g of AlCl3

STEP 1 Given: 125 mL (0.125 L) of solution

0.150 M AlCl3 solution

Need: g of AlCl3

STEP 2 Plan:

L of solution moles of AlCl3 g of AlCl3

STEP 3Write equalities and conversion factors:

1 mole of AlCl3 = 133.5 g of AlCl3

1 mole AlCl3 and 133.5 g AlCl3

133.5 g AlCl3 1 mole AlCl3

Solution (continued)

STEP 3(continued)

1 L of KCl = 0.150 mole of AlCl3

1 L and 0.150 mole AlCl3

0.150 mole AlCl3 1 L

STEP 4 Set up problem:

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

1 L 1 mole AlCl3

Learning Check

How many milliliters of 2.00 M HNO3 contain24.0 g of HNO3?

1) 12.0 mL of 2.00 M HNO3

2) 83.3 mL of 2.00 M HNO3

3) 190 mL of 2.00 M HNO3

Solution

3) 190 mL of HNO3

STEP 1 Given: 24.0 g of HNO3

2.00 M HNO3 solution

Need: mL of HNO3 solution

STEP 2 Plan:

g of HNO3 moles of HNO3 L of HNO3 solution

STEP 3Write equalities and conversion factors:

1 mole of = 63.0 g of HNO3

1 mole HNO3 and 63.0 g HNO3

63.0 g HNO3 1 mole HNO3

Solution (continued)

STEP 4 Set up problem to calculate volume, in mL, of

HNO3:

24.0 g HNO3 x 1 mole HNO3 x 1000 mL HNO3

63.0 g HNO3 2.00 moles HNO3

= 190 mL of a 2.00 M HNO3 solution

Dilution

In a dilution,

• volume increases
• concentration of solute decreases
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

Concentrated Diluted

solution solution

For molarity

M1V1 = M2V2

Concentrated Diluted

solution solution

Example of Dilution Calculations Using Percent Concentration

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?

STEP 1 Prepare a table:

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

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

STEP 2 Solve dilution expression for the unknown

C1V1 = C2V2 V2 =V1C1

C2

STEP 3Set up the problem using known quantities:

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

C2 2.00%

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?

Solution

What is the percent (m/v) of a solution prepared

by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?

STEP 1 Prepare a table:

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

C2 = ?V2 = 60.0 mL

STEP 2 Solve the dilution expression for the unknown: C1V1 = C2V2 C2 =C1V1

V2

STEP 3Set up the problem using known quantities:

C2 =C1V1 = (10.0 mL)(9.00%) = 1.50% (m/v)

V2 60.0 mL

Example of Dilution Calculations Using Molarity

What is the molarity (M) of a solution prepared

by diluting 0.180 L of 0.600 M HNO3 to 0.540 L?

STEP 1 Prepare a table:

M1 = 0.600 M V1 = 0.180 L

M2= ?V2 = 0.540 L

STEP 2 Solve the dilution expression for the unknown: M1V1 = M2V2

STEP 3Set up the problem using known quantities:

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

V2 0.540 L

Learning Check

What is the final volume (mL) of 15.0 mL of a 1.80 M

KOH diluted to give a 0.300 M solution?

1) 27.0 mL of a 1.80 M KOH

2) 60.0 mL of a 1.80 M KOH

3) 90.0 mL of a 1.80 M KOH

Solution

What is the final volume (mL) of 15.0 mL of a 1.80 M

KOH diluted to give a 0.300 M solution?

STEP 1 Prepare a table:

M1 = 1.80 M V1 = 15.0 mL

M2 = 0.300 MV2 = ?

STEP 2 Solve the dilution expression for the unknown: M1V1 = M2V2

STEP 3Set up the problem using known quantities:

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

M2 0.300 M

Molarity in Chemical Reactions

In a chemical reaction,

• the volume and molarity of a solution are used to determine the moles of a reactant or product

molarity ( mole ) x volume (L) = moles

1 L

• if molarity (mole/L) and moles are given, the volume (L) can be determined

moles x 1 L = volume (L)

moles

Using Molarity of Reactants

How many mL of 3.00 M HCl are needed to completely

react with 4.85 g of CaCO3?

2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)

STEP 1 Given: 3.00 M HCl; 4.85 g of CaCO3

Need: volume of HCl in mL

STEP 2Write a plan:

g of CaCO3 moles of CaCO3 moles of HCl mL of HCl

STEP 3Write equalities and conversion factors:

1 mole of CaCO3 = 100.1 g

1 mole CaCO3 and 100.1 g CaCO3

100.1 g CaCO3 1 mole CaCO3

Using Molarity of Reactants (continued)

STEP 3(continued)

1 mole of CaCO3 = 2 moles of HCl

1 mole of CaCO3 and 2 moles of HCl

2 moles of HCl 1 mole of CaCO3

1000 mL of HCl = 3.00 moles of HCl

1000 mL of HCland 3.00 moles of HCl

3.00 moles of HCl 1000 mL of HCl

STEP 4 Set up problem to calculate mL of HCl:

4.85 g CaCO3 x 1 mole CaCO3 x 2 moles HCl x 1000 mL HCl

100.1 g CaCO3 1 mole CaCO3 3.00 moles HCl

= 32.3 mL of HCl required

Learning Check

If 22.8 mL of 0.100 M MgCl2 is needed to completely react 15.0 mL of AgNO3 solution, what is the molarity of the AgNO3 solution?

MgCl2(aq) + 2AgNO3(aq) 2AgCl(s) + Mg(NO3)2(aq)

1) 0.0760 M

2) 0.152 M

3) 0.304 M

Solution

3) 0.304 M AgNO3

STEP 1 Given: 22.8 mL (0.228 L) of 0.100 M MgCl2 Need: molarity of AgNO3

STEP 2Write a plan to calculate molarity:

mL of MgCl2 moles of MgCl2 moles of AgNO3 molarity of AgNO3

STEP 3Write equalities and conversion factors:

0.100 mole of MgCl2 = 1 L of MgCl2

0.100 mole MgCl2 and 1 L MgCl2 1 L MgCl2 0.100 mole MgCl2

Solution (continued)

STEP 3 (continued)

1 mole of MgCl2 = 2 moles AgNO3

2 moles AgNO3 and 1 mole MgCl2

1 mole MgCl2 2 moles AgNO3

STEP 4 Set up problem to calculate molarity of AgNO3:

0.0228 L x 0.100 mole MgCl2 x 2 moles AgNO3 x 1____

1 L 1 mole MgCl2 0.0150 L

= 0.304 mole/L = 0.304 M AgNO3

Learning Check

How many liters of H2 gas at STP are produced when125 mL of 6.00 M HCl reacts with sufficient Zn?

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

1) 4.20 L of H2

2) 8.40 L of H2

3) 16.8 L of H2

Solution

2) 8.40 L of H2 gas

STEP 1 Given: 125 mL (0.125 L) of 6.00 M HCl Need: liters of H2 at STP

STEP 2Write a plan to calculate liters of H2:

L of HCl moles of HCl moles of H2 liters of H2

STEP 3Write equalities and conversion factors:

6.00 moles of HCl = 1 L of HCl

6.00 moles HCland 1 L HCl

1 L HCl 6.00 moles HCl

22.4 L of H2 = 1 mole of H2 at STP

22.4 L H2 and 1 mole H2

1 mole H2 22.4 L H2

Solution (continued)

STEP 3 (continued)

2 moles HCl = 1 mole H2

2 moles HCland 1 mole H2

1 mole H2 2 moles HCl

STEP 4 Set up problem to calculate liters of H2:

0.125 L x 6.00 moles HCl x 1 mole H2 x 22.4 L

1 L 2 moles HCl 1 mole H2

= 8.40 L of H2 gas