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MOLARITY – Ch 13, p. 412MOLARITY – Ch 13, p. 412
Quantifies the concentration of a solution.
Molarity (M) = mol solute = n = mol
volume solution V L
Read as “moles solute per liter of solution”
UnitsVariables
MOLARITY – Ch 13MOLARITY – Ch 13Example 1: A bottle labeled “0.2 M
NaCl” reads as
“0.2 molar sodium chloride solution.”
MOLARITY – Ch 13MOLARITY – Ch 13
Example 1: A bottle labeled “0.2 M NaCl” reads as
“0.2 molar sodium chloride solution.”
This means 1 L of the solution contains 0.2 moles NaCl.
MOLARITY – Ch 13MOLARITY – Ch 13
Example 2: Calculate the molarity of a solution prepared by dissolving 5.844 g NaCl in enough water to make a 200.0 mL solution.
Example 2: Calculate the molarity of a solution prepared by dissolving 5.844 g NaCl in enough water to make a 200.0 mL solution.
Given:
Example 2: Calculate the molarity of a solution prepared by dissolving 5.844 g NaCl in enough water to make a 200.0 mL solution.
Given: mass = 5.844 g
Example 2: Calculate the molarity of a solution prepared by dissolving 5.844 g NaCl in enough water to make a 200.0 mL solution.
Given: mass = 5.844 g V = 200.0 mL
Example 2: Calculate the molarity of a solution prepared by dissolving 5.844 g NaCl in enough water to make a 200.0 mL solution.
Given: mass = 5.844 g V = 200.0 mL
Unknown:
Example 2: Calculate the molarity of a solution prepared by dissolving 5.844 g NaCl in enough water to make a 200.0 mL solution.
Given: mass = 5.844 g V = 200.0 mL
Unknown: M
Example 2: Calculate the molarity of a solution prepared by dissolving 5.844 g NaCl in enough water to make a 200.0 mL solution.
Given: mass = 5.844 g V = 200.0 mL
Unknown: M
Need:
MOLARITY – Ch 13MOLARITY – Ch 13
Example 2: Calculate the molarity of a solution prepared by dissolving 5.844 g NaCl in enough water to make a 200.0 mL solution.
Given: mass = 5.844 g V = 200.0 mL
Unknown: M
Need: molar mass NaCl
MOLARITY – Ch 13MOLARITY – Ch 13
Example 2:Given: mass = 5.844 g
V = 200.0 mL = 0.2000 LUnknown: MNeed: molar mass NaCl 58.443 g/mol
M = n = 5.844 g V
MOLARITY – Ch 13MOLARITY – Ch 13Example 2:
Given: mass = 5.844 g V = 200.0 mL =
0.2000 LUnknown: MNeed: molar mass NaCl
M = n = 5.844 g mol V 58.443 g
MOLARITY – Ch 13MOLARITY – Ch 13
Example 2:Given: mass = 5.844 g
V = 200.0 mL = 0.2000 LUnknown: MNeed: molar mass NaCl
M = n = 5.844 g mol V 58.443 g
MOLARITY – Ch 13MOLARITY – Ch 13
Example 2:Given: mass = 5.844 g
V = 200.0 mL = 0.2000 LUnknown: MNeed: molar mass NaCl
M = n = 5.844 g mol V 0.2000 L 58.443 g
MOLARITY – Ch 13MOLARITY – Ch 13
Example 2:Given: mass = 5.844 g
V = 200.0 mL = 0.2000 LUnknown: MNeed: molar mass NaCl
M = n = 5.844 g mol 0.49997 M
V 0.2000 L 58.443 g
MOLARITY – Ch 13MOLARITY – Ch 13
Example 2:Given: mass = 5.844 g
V = 200.0 mL = 0.2000 LUnknown: MNeed: molar mass NaCl
M = n = 5.844 g mol 0.5000 M
V 0.2000 L 58.443 g
MOLARITY – Ch 13MOLARITY – Ch 13
Example 3: Calculate the moles of NaOH present in 5000.0 mL of 0.250 M NaOH.
MOLARITY – Ch 13MOLARITY – Ch 13
Example 3: Calculate the moles of NaOH present in 5000.0 mL of 0.250 M NaOH.
Given:
MOLARITY – Ch 13MOLARITY – Ch 13
Example 3: Calculate the moles of NaOH present in 5000.0 mL of 0.250 M NaOH.
Given: V = 5000.0 mL
MOLARITY – Ch 13MOLARITY – Ch 13
Example 3: Calculate the moles of NaOH present in 5000.0 mL of 0.250 M NaOH.
Given: V = 5000.0 mLmolarity = 0.250 M
MOLARITY – Ch 13MOLARITY – Ch 13
Example 3: Calculate the moles of NaOH present in 5000.0 mL of 0.250 M NaOH.
Given: V = 5000.0 mLmolarity = 0.250 M
Unknown: n, mol NaOH
MOLARITY – Ch 13MOLARITY – Ch 13
Example 3: Calculate the moles of NaOH present in 5000.0 mL of 0.250 M NaOH.
Given: V = 5000.0 mLmolarity = 0.250 M
Unknown: n, mol NaOH
Need:
MOLARITY – Ch 13MOLARITY – Ch 13Example 3: Calculate the moles of
NaOH present in 5000.0 mL of 0.250 M NaOH.
Given: V = 5000.0 mLmolarity = 0.250 M
Unknown: n, mol NaOH
Need: M = n V
MOLARITY – Ch 13MOLARITY – Ch 13Example 3: Calculate the moles of
NaOH present in 5000.0 mL of 0.250 M NaOH.
Given: V = 5000.0 mLmolarity = 0.250 M
Unknown: n, mol NaOH
Need: M = n V
MOLARITY – Ch 13MOLARITY – Ch 13Example 3: Calculate the moles of
NaOH present in 5000.0 mL of 0.250 M NaOH.
Given: V = 5000.0 mLmolarity = 0.250 M
Unknown: n, mol NaOH
Need: M = n n = M•VV
MOLARITY – Ch 13MOLARITY – Ch 13
Example 3:Given: V = 5000.0 mL
molarity = 0.250 MUnknown: n, mol NaOHNeed: M = n n = M•V
V 0.250 M
MOLARITY – Ch 13MOLARITY – Ch 13
Example 3:Given: V = 5000.0 mL
molarity = 0.250 MUnknown: n, mol NaOHNeed: M = n n = M•V
V 0.250 mol
L
MOLARITY – Ch 13MOLARITY – Ch 13
Example 3:Given: V = 5000.0 mL = 5.0000 L
molarity = 0.250 MUnknown: n, mol NaOHNeed: M = n n = M•V
V 0.250 mol
L
MOLARITY – Ch 13MOLARITY – Ch 13
Example 3:Given: V = 5000.0 mL = 5.0000 L
molarity = 0.250 MUnknown: n, mol NaOHNeed: M = n n = M•V
V 0.250 mol 5.0000 L
L
MOLARITY – Ch 13MOLARITY – Ch 13
Example 3:Given: V = 5000.0 mL = 5.0000 L
molarity = 0.250 MUnknown: n, mol NaOHNeed: M = n n = M•V
V 0.250 mol 5.0000 L
L
MOLARITY – Ch 13MOLARITY – Ch 13
Example 3:Given: V = 5000.0 mL = 5.0000 L
molarity = 0.250 MUnknown: n, mol NaOHNeed: M = n n = M•V
V 0.250 mol 5.0000 L 1.25
mol L
MOLARITY – Ch 13MOLARITY – Ch 13
Example 4: Calculate the volume of 0.50 M HCl that would contain 2.0 mol HCl.
YOU TRY IT!
MOLARITY – Ch 13MOLARITY – Ch 13
Example 4:Given: n = 2.0 mol HCl
molarity = 0.50 MUnknown: VNeed: M = n V = n
V M2.0 mol L 4.0 L
0.50 mol
MOLARITY – Ch 13MOLARITY – Ch 13
Example 5: How would you prepare 100 mL of 0.50 M HCl from concentrated 12 M HCl?
YOU TRY IT!
MOLARITY – Ch 13MOLARITY – Ch 13
Example 5: How would you prepare 100 mL of 0.50 M HCl from concentrated 12 M HCl?
Use C1V1 = C2V2
MOLARITY – Ch 13MOLARITY – Ch 13
Example 5: How would you prepare 100 mL of 0.50 M HCl from concentrated 12 M HCl?
Use C1V1 = C2V2
Given: C1 = 0.50 M V1 = 100mL
C2 = 12 M
MOLARITY – Ch 13MOLARITY – Ch 13
Example 5: How would you prepare 100 mL of 0.50 M HCl from concentrated 12 M HCl?
Use C1V1 = C2V2
Given: C1 = 0.50 M V1 = 100mL
C2 = 12
Unknown:
MOLARITY – Ch 13MOLARITY – Ch 13
Example 5: How would you prepare 100 mL of 0.50 M HCl from concentrated 12 M HCl?
Use C1V1 = C2V2
Given: C1 = 0.50 M V1 = 100mL
C2 = 12
Unknown: V2 =
MOLARITY – Ch 13MOLARITY – Ch 13
Example 5: How would you prepare 100 mL of 0.50 M HCl from concentrated 12 M HCl?
Use C1V1 = C2V2 C1V1 = V2
C2
Given: C1 = 0.50 M V1 = 100mL
C2 = 12
Unknown: V2 =
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