(a)
Interpretation:
The new molarity of the given solution diluted with 150 mL of water is to be calculated.
Concept Introduction:
The molarity of a solution is defined as the number of moles of solute dissolved in one liter of the solution. The formula for molarity is given as:
M=nV
Where,
- n represents the number of moles of the solute.
- V represents the volume of the solution.
Answer to Problem 122AP
The new molarity of the 0.200 M HBr solution on dilution is 0.091 M.
Explanation of Solution
The molarity of the given HBr solution is 0.200 M.
The volume of 0.200 M HBr solution is 125 mL.
The volume of water added in the solution is 150 mL.
The total new volume of the solution is given as:
Vf=Vi+Vw
Where,
- Vi represents the initial volume of the solution.
- Vw represents the volume of water added in the solution.
Substitute the value of Vi and Vw in the above equation.
Vf=125 mL+150 mL=275 mL
The relation between the initial and final volume of a solution is given as:
MfVf=MiVi
Where,
- Mf represents the final molarity of the solution.
- Vf represents the final volume of the solution.
- Mi represents the initial molarity of the solution.
- Vi represents the initial volume of the solution.
Rearrange the above equation for the value of Mf.
MfVf=MiVi
Substitute the value of Vf, Mi and Vi in the above equation.
Mf=(0.200 M)(125 mL)275 mL=0.091 M
Therefore, the new molarity of the 0.200 M HBr solution on dilution is 0.091 M.
(b)
Interpretation:
The new molarity of the given solution diluted with 150 mL of water is to be calculated.
Concept Introduction:
The molarity of a solution is defined as the number of moles of solute dissolved in one liter of the solution. The formula for molarity is given as:
M=nV
Where,
- n represents the number of moles of the solute.
- V represents the volume of the solution.
Answer to Problem 122AP
The new molarity of the 0.250 M Ca(C2H3O2)2 solution on dilution is 0.127 M.
Explanation of Solution
The molarity of the given Ca(C2H3O2)2 solution is 0.250 M.
The volume of 0.250 M Ca(C2H3O2)2 solution is 155 mL.
The volume of water added in the solution is 150 mL.
The total new volume of the solution is given as:
Vf=Vi+Vw
Where,
- Vi represents the initial volume of the solution.
- Vw represents the volume of water added in the solution.
Substitute the value of Vi and Vw in the above equation.
Vf=155 mL+150 mL=305 mL
The relation between the initial and final volume of a solution is given as:
MfVf=MiVi
Where,
- Mf represents the final molarity of the solution.
- Vf represents the final volume of the solution.
- Mi represents the initial molarity of the solution.
- Vi represents the initial volume of the solution.
Rearrange the above equation for the value of Mf.
MfVf=MiVi
Substitute the value of Vf, Mi and Vi in the above equation.
Mf=(0.250 M)(155 mL)305 mL=0.127 M
Therefore, the new molarity of the 0.250 M Ca(C2H3O2)2 solution on dilution is 0.127 M.
(c)
Interpretation:
The new molarity of the given solution diluted with 150 mL of water is to be calculated.
Concept Introduction:
The molarity of a solution is defined as the number of moles of solute dissolved in one liter of the solution. The formula for molarity is given as:
M=nV
Where,
- n represents the number of moles of the solute.
- V represents the volume of the solution.
Answer to Problem 122AP
The new molarity of the 0.250 M H3PO4 solution on dilution is 0.192 M.
Explanation of Solution
The molarity of the given H3PO4 solution is 0.250 M.
The volume of 0.250 M H3PO4 solution is 0.500 L.
The volume of water added in the solution is 150 mL.
The total new volume of the solution is given as:
Vf=Vi+Vw
Where,
- Vi represents the initial volume of the solution.
- Vw represents the volume of water added in the solution.
Substitute the value of Vi and Vw in the above equation.
Vf=(0.500 L)(1000 mL1 L)+150 mL=650 mL
The relation between the initial and final volume of a solution is given as:
MfVf=MiVi
Where,
- Mf represents the final molarity of the solution.
- Vf represents the final volume of the solution.
- Mi represents the initial molarity of the solution.
- Vi represents the initial volume of the solution.
Rearrange the above equation for the value of Mf.
MfVf=MiVi
Substitute the value of Vf, Mi and Vi in the above equation.
Mf=(0.250 M)(0.500 L)(1000 mL1 L)650 mL=0.192 M
Therefore, the new molarity of the 0.250 M H3PO4 solution on dilution is 0.192 M.
(d)
Interpretation:
The new molarity of the given solution diluted with 150 mL of water is to be calculated.
Concept Introduction:
The molarity of a solution is defined as the number of moles of solute dissolved in one liter of the solution. The formula for molarity is given as:
M=nV
Where,
- n represents the number of moles of the solute.
- V represents the volume of the solution.
Answer to Problem 122AP
The new molarity of the 18.0 M H2SO4 solution on dilution is 1.636 M.
Explanation of Solution
The molarity of the given H2SO4 solution is 18.0 M.
The volume of 18.0 M H2SO4 solution is 15 mL.
The volume of water added in the solution is 150 mL.
The total new volume of the solution is given as:
Vf=Vi+Vw
Where,
- Vi represents the initial volume of the solution.
- Vw represents the volume of water added in the solution.
Substitute the value of Vi and Vw in the above equation.
Vf=15 mL+150 mL=165 mL
The relation between the initial and final volume of a solution is given as:
MfVf=MiVi
Where,
- Mf represents the final molarity of the solution.
- Vf represents the final volume of the solution.
- Mi represents the initial molarity of the solution.
- Vi represents the initial volume of the solution.
Rearrange the above equation for the value of Mf.
MfVf=MiVi
Substitute the value of Vf, Mi and Vi in the above equation.
Mf=(18.0 M)(15 mL)165 mL=1.636 M
Therefore, the new molarity of the 18.0 M H2SO4 solution on dilution is 1.636 M.
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Chapter 15 Solutions
EBK INTRO.CHEMISTRY (NASTA EDITION)
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