(a)
Interpretation:
The volume of water should be added to yield the given solution has to be determined.
Concept Introduction:
Dilution equation:
Dilution equation is given by,
(a)
Answer to Problem 8.67EP
The volume of water should be added to yield the given solution is 1450 mL.
Explanation of Solution
Given information:
Volume of solution = 50.0 mL
Initial concentration = 3.00 M
Required concentration = 0.100 M
Final volume of solution = ?
Apply dilution equation to calculate the volume of water should be added to yield the given solution,
Now subtract the original volume (Vd) from the final volume (Vs)
Therefore, the volume of water should be added to yield the given solution is 1450 mL.
(b)
Interpretation:
The volume of water should be added to yield the given solution has to be determined.
Concept Introduction:
Dilution equation:
Dilution equation is given by,
(b)
Answer to Problem 8.67EP
The volume of water should be added to yield the given solution is 18.0 mL.
Explanation of Solution
Given information:
Volume of solution = 2.00 mL
Initial concentration = 1.00 M
Required concentration = 0.100 M
Final volume of solution = ?
Apply dilution equation to calculate the volume of water should be added to yield the given solution,
Now subtract the original volume (Vd) from the final volume (Vs)
Therefore, the volume of water should be added to yield the given solution is 18.0 mL.
(c)
Interpretation:
The volume of water should be added to yield the given solution has to be determined.
Concept Introduction:
Dilution equation:
Dilution equation is given by,
(c)
Answer to Problem 8.67EP
The volume of water should be added to yield the given solution is 85550 mL.
Explanation of Solution
Given information:
Volume of solution = 1450 mL
Initial concentration = 6.00 M
Required concentration = 0.100 M
Final volume of solution = ?
Apply dilution equation to calculate the volume of water should be added to yield the given solution,
Now subtract the original volume (Vd) from the final volume (Vs)
Therefore, the volume of water should be added to yield the given solution is 85550 mL.
(d)
Interpretation:
The volume of water should be added to yield the given solution has to be determined.
Concept Introduction:
Dilution equation:
Dilution equation is given by,
(d)
Answer to Problem 8.67EP
The volume of water should be added to yield the given solution is 7.5 mL.
Explanation of Solution
Given information:
Volume of solution = 75.0 mL
Initial concentration = 0.110 M
Required concentration = 0.100 M
Final volume of solution = ?
Apply dilution equation to calculate the volume of water should be added to yield the given solution,
Now subtract the original volume (Vd) from the final volume (Vs)
Therefore, the volume of water should be added to yield the given solution is 7.5 mL.
Want to see more full solutions like this?
Chapter 8 Solutions
General, Organic, and Biological Chemistry
- 3.63 How many moles of solute are present in each of these solutions? (a) 48.0 mL of 3.4 M H2SO4. (b) 1.43 mL of 5.8 M KNO3. (c) 321 L of 0.034M NH3 (d) 1.9 × 10-3 L of 1.4 × 10-5 M NaFarrow_forward34. For each of the following solutions, the number of moles of solute is given, followed by the total volume of the solution prepared. Calculate the molarity of each solution. a. 0.754 mol KNO; 225 mL b. 0.0105 in of CaCl; 10.2 mL c. 3.15 mol NaCl; 5.00 L d. 0.499 mol NaBr; 100. mLarrow_forwardCalcium carbonate, CaCO3, can be obtained in a very pure state. Standard solutions of calcium ion are usually prepared by dissolving calcium carbonate in acid. What mass of CaCO3 should be taken to prepare 500. mL of 0.0200 M calcium ion solution?arrow_forward
- A large beaker contains 1.50 L of a 2.00 M iron(III) chloride solution. How many moles of iron ions are in the solution? How many moles of chloride ions are in the solution? You now add 0.500 L of a 4.00 M lead(II) nitrate solution to the beaker. Determine the mass of solid product formed (in grams).arrow_forward3.65 Determine the final molarity for the following dilutions. (a) 24.5 mL of 3.0 M solution diluted to 100.0 mL (b) 15.3 mL of 4.22 M solution diluted to 1.00 L (c) 1.45 mL of 0.034 M solution diluted to 10.0 mL (d) 2.35 L of 12.5 M solution diluted to 100.0 Larrow_forward
- General, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage LearningIntroductory Chemistry: A FoundationChemistryISBN:9781337399425Author:Steven S. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistry for Engineering StudentsChemistryISBN:9781337398909Author:Lawrence S. Brown, Tom HolmePublisher:Cengage Learning
- Living By Chemistry: First Edition TextbookChemistryISBN:9781559539418Author:Angelica StacyPublisher:MAC HIGHERChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning