14 Solutions Problem Set
SOLUTIONS | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: Environmental engineers typically do calculations involving the mixture of water samples and various chemical additives to measure certain physical characteristics. Given water is added to 25 mL of a 1.25 M πΎ2 ππ4 stock solution, the amount of water required to result in a 500 mL solution having a potassium ion concentration of 0.500 M is most close to: A. 250 ππΏ B. 350 ππΏ C. 375 ππΏ D. 475 ππΏ
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SOLUTION 1: The TOPIC of SOLUTION CONCENTRATIONS is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.
In this problem, we are looking to SOLVE for the AMOUNT of WATER that should be ADDED to in a CONCENTRATION of POTASSIUM IONS (πΎ + ) of 0.500 M. As we are GIVEN the CONCENTRATIONS and VOLUME of each SUBSTANCE, we can USE a STOICHIOMETRIC RATIO to determine the CONCENTRATION of POTASSIUM IONS (πΎ + ). However, before can SOLVE for the CONCENTRATION of POTASSIUM IONS (πΎ + ), we need to CALCULATE the CONCENTRATION of the STOCK SOLUTION in the overall SOLUTION. As we are working with STOCK SOLUTION of πΎ2 ππ4, we know that the MAJORITY COMPONENT of the SOLUTION will be the WATER that is MIXED in to DILUTE the STOCK SOLUTION. Therefore, the STOCK SOLUTION of πΎ2 ππ4 is the SOLUTE, and the WATER being ADDED is the SOLVENT.
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The term CONCENTRATION is used to INDICATE the AMOUNT of SOLUTE dissolved in a given quantity of SOLVENT or SOLUTION. Brackets β[ ]β are commonly used to indicate that a value REPRESENTS the CONCENTRATION of a SUBSTANCE, and have associated UNITS of MOLES of SOLUTE per LITER of SOLUTION. The CONCENTRATION is defined as the PERCENTAGE of the SOLUTE present within a given VOLUME of SOLUTION.
πΆππππππ‘πππ‘πππ =
ππππ ππ ππππ’π‘π πΏππ‘ππ ππ ππππ’π‘πππ
The CONCENTRATION represents the PHYSICAL PARAMETER of VOLUME and is commonly expressed as the MOLARITY or MOLALITY of a SOLUTION. When the CONCENTRATION is expressed on a VOLUMETRIC SCALE such that the AMOUNT of SOLUTE contained in a given VOLUME of SOLUTION is EQUAL to the PRODUCT of the VOLUME, the CONCENTRATION is given as: π΄πππ’ππ‘ ππ ππππ’π‘π = ππππ’ππ Γ πΆππππππ‘πππ‘πππ We should NOTE that the VOLUME is given in UNITS of MILLILITERS, so we need to take note of making sure all of our UNITS are CONSISTENT.
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The TABLE of METRIC PREFIXES can be referenced under the SUBJECT of UNITS on page 1 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Using the METRIX PREFIX for milli-, the CONVERSION FACTOR for converting LITERS (L) to MILLILITERS (mL) is given as: 1 ππΏ = 1 Γ 10β3 πΏ As we are told that 25 mL of a 1.25 M πΎ2 ππ4 stock solution is being used, we can SOLVE the AMOUNT of SOLUTE in the SOLUTION as: 1 Γ 10β3 πΏ 1.25 πππ πΎ2 ππ4 π΄πππ’ππ‘ ππ ππππ’π‘π = 25 ππΏ Γ Γ = 0.03125 πππ πΎ2 ππ4 1 ππΏ 1πΏ Now that we know the AMOUNT of STOCK SOLUTION we are working with, we need to take it one step further to CALCULATED the amount of POTASSIUM IONS (πΎ + ) in the SOLUTION. We can then use a MOLAR RATIO to COMPARE the number of MOLES of POTASSIUM (πΎ + ) in a MOLECULE of πΎ2 ππ4 . As there are 2 POTASSIUMS IONS (πΎ + ) in every MOLECULE of πΎ2 ππ4, we can CALCULATE the CONCENTRATIONS of POTASSIUMS IONS [πΎ + ] in the SOLUTE as:
[πΎ + ]
2 ππππ πΎ + = 0.03125 ππΏ πΎ2 ππ4 Γ = 0.0625 πππ πΎ + 1 πππ πΎ2 ππ4 Made with
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Recall the CONCENTRATION is defined as the PERCENTAGE of the SOLUTE present within a given VOLUME of SOLUTION. Based on the DEFINED concentration of [πΎ + ] = 0.5 π from the PROBLEM STATEMENT, we can RELATE the NUMBER of MOLES of POTASSIUM IONS to an VOLUME representing the MOLES of SOLUTE:
0.5 π
[πΎ + ]
ππππ πΎ + ππ π πππ’π‘π = 1 πΏ ππ π πππ£πππ‘
Now that we have defined CONCENTRATION of POTASSIUM IONS in the SOLUTE as [πΎ + ] = 0.0625 πππ πΎ + , we can PLUG in the GIVEN potassium ion concentration level of 0.5 M to calculate the MOLS of SOLUTE in TERMS of VOLUME as:
0.0625 πππ πΎ + Γ
1 πΏ ππ π πππ£πππ‘ = 0.125 πΏ = 125 ππΏ 0.500 πππ πΎ +
As we are told the VOLUME of the SOLUTION is 500 mL, we can subtract the VOLUME of the POTASSIUM IONS to calculate the VOLUME of water ADDED as: 500 ππΏ β 125 ππ = 375 ππΏ
Therefore, the correct answer choice is C. 375 mL
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PROBLEM 2: Given a sample solution with a volume of 375 mL that is measured to contain 0.278 moles of solute, the molarity of the solution is most close to: A. 0.218 B. 0.378 C. 0.741 D. 0.812
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SOLUTION 2: The TOPIC of MOLARITY OF SOLUTIONS can be referenced under SUBJECT of CHEMISTRY on page 54 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. In this problem, we are looking to SOLVE for the MOLARITY of a SOLUTION given the MOLES of SOLUTE and VOLUME of SOLUTION. The MOLARITY (M) of a SOLUTION is the NUMBER of GRAM MOLES of a substance DISSOLVED in a LITER VOLUME of SOLUTION.
πππππππ‘π¦ (π) =
ππππ ππ ππππ’π‘π πΏππ‘ππ ππ ππππ’π‘πππ
We are TOLD that the VOLUME of the SOLUTION sample is 375 mL. As the UNITS of MOLARITY are in LITERS (L), we need to CONVERT the VOLUME from milliliters (mL) to liters (L). Doing so, we find the VOLUME in UNITS of milliliters (mL) is: 1 Γ 10β3 πΏ ππππ’ππ = 375 ππΏ Γ = 0.375 πΏ 1 ππΏ We are also TOLD that the MOLES of SOLUTE in the SOLUTION is 0.278 moles, which we can PLUG directly into the FORMULA for MOLARITY. Made with
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As we now have the VOLUME in liters, and are given the MOLES of SOLUTE, we can simply PLUG and CHUG these VALUES in the FORMULA for MOLARITY and find:
πππππππ‘π¦ (π) =
ππππ ππ ππππ’π‘π 0.278 ππππ πππ = = 0.741 = 0.741 π πΏππ‘ππ ππ ππππ’π‘πππ 0.375 πΏ πΏ
Therefore, the correct answer choice is C. 0.741
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PROBLEM 3: Given 14.38 moles of potassium fluoride is dissolved to make 13 mL of solution, the molarity of the solution is most close to: A. 854 B. 1106 C. 1352 D. 2357
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SOLUTION 3: The TOPIC of MOLARITY OF SOLUTIONS can be referenced under SUBJECT of CHEMISTRY on page 54 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. In this problem, we are looking to SOLVE for the MOLARITY of a SOLUTION given the MOLES of SOLUTE and VOLUME of SOLUTION. The MOLARITY (M) of a SOLUTION is the NUMBER of GRAM MOLES of a substance DISSOLVED in a LITER VOLUME of SOLUTION.
πππππππ‘π¦ (π) =
ππππ ππ ππππ’π‘π πΏππ‘ππ ππ ππππ’π‘πππ
We are TOLD that the VOLUME of the SOLUTION sample is 13 mL. As the UNITS of MOLARITY are in LITERS (L), we need to CONVERT the VOLUME from milliliters (mL) to liters (L). Doing so, we find the VOLUME in UNITS of milliliters (mL) is: 1 Γ 10β3 πΏ ππππ’ππ = 13 ππΏ Γ = 0.013 πΏ 1 ππΏ We are also TOLD that the MOLES of SOLUTE in the SOLUTION is 14.38 moles, which we can PLUG directly into the FORMULA for MOLARITY. Made with
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As we now have the VOLUME in liters, and are given the MOLES of SOLUTE, we can simply PLUG and CHUG these VALUES in the FORMULA for MOLARITY and find:
πππππππ‘π¦ (π) =
ππππ ππ ππππ’π‘π 14.38 ππππ πΎπΉ πππ = = 1106.15 = 1106 π πΏππ‘ππ ππ ππππ’π‘πππ 0.013 πΏ πΏ
Therefore, the correct answer choice is B. 1106
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PROBLEM 4: If 25 mL of water is added to 375 mL of a 0.65 M peroxide solution, the molarity of the diluted solution is most close to: A. 0.12 B. 0.28 C. 0.53 D. 0.61
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SOLUTION 4: The TOPIC of DILUTION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. In this problem, we are GIVEN the INITIAL CONCENTRATION and VOLUME of a SOLUTION, and then asked to SOLVE for the FINAL CONCENTRATION of the SOLUTION, once some DEFINED AMOUNT of SOLVENT is ADDED to the SOLUTION. When working with SOLUTIONS in CHEMISTRY, it is common to need to CHANGE the CONCENTRATION of a SOLUTION by changing the AMOUNT of SOLVE. A DILUTION is the ADDITION of SOLVENT to a SOLUTION, which DECREASES the CONCENTRATION of the SOLUTE in the SOLUTION. DILUTION alters the MOLARITY or CONCENTRATION of the SOLUTION such that the NUMBER of MOLES of SOLUTE can be expressed as: πππππ ππ ππππ’π‘π = (πππππππ‘π¦)(πΏππ‘πππ ππ ππππ’π‘πππ) = (π)(π) The FORMULA for DILUTION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. Made with
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The standard EQUATION for DILUTION compares the INITIAL and FINAL CONDITIONS of the SOLUTION, as shown by the expression: (π1 )(π1 ) = (π2 )(π2 ) Where: β’ π is the MOLARITY or CONCENTRATION given in units of moles per liter (πππ/πΏ) β’ π is the VOLUME given in units of liters (πΏ) β’ The subscript 1 refers to BEFORE DILUTION β’ The subscript 2 refers to AFTER DILUTION
We are told that the INITIAL VOLUME of the SOLUTION is π1 = 375 ππΏ and the INITIAL MOLARITY of the SOLUTION is π1 = 0.65 π. For the FINAL CONDITION, we are told that the VOLUME is to INCREASE by 25 mL, and then asked to SOLVE for the FINAL CONCENTRATION of the SOLUTION. We can CALCULATE the FINAL VOLUME (π2 ), as the SUM of the INITIAL VOLUME plus the ADDITION of 25 mL of SOLVENT. π2 = 375 ππΏ + 25 ππΏ = 400 ππΏ
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Plugging in the INITIAL VOLUME, INITIAL CONCENTRATION, and FINAL VOLUME into the FORMULA for DILUTION, we can re-write the EQUATION for DILUTION as: (0.65 π)(375 ππΏ) = π2 (400 ππΏ) SOLVING for the FINAL CONCENTRATION of the SOLUTION, we find:
π2 =
(0.65 π)(375 ππΏ) = 0.609 π 400 ππΏ
Therefore, the correct answer choice is D. 0.61
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PROBLEM 5: The amount of 0.075 M HF solution that can be made by diluting 525 mL of 5.5 M HF in units of liters is most close to: A. 37.0 B. 38.5 C. 42.5 D. 44.0
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SOLUTION 5: The TOPIC of DILUTION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. In this problem, we are GIVEN the INITIAL CONCENTRATION and VOLUME of a SOLUTION, and then asked to SOLVE for the FINAL VOLUME of the SOLUTION, GIVEN the FINAL CONCENTRATION of the SOLUTION. When working with SOLUTIONS in CHEMISTRY, it is common to need to CHANGE the CONCENTRATION of a SOLUTION by changing the AMOUNT of SOLVE. A DILUTION is the ADDITION of SOLVENT to a SOLUTION, which DECREASES the CONCENTRATION of the SOLUTE in the SOLUTION. DILUTION alters the MOLARITY or CONCENTRATION of the SOLUTION such that the NUMBER of MOLES of SOLUTE can be expressed as: πππππ ππ ππππ’π‘π = (πππππππ‘π¦)(πΏππ‘πππ ππ ππππ’π‘πππ) = (π)(π) The FORMULA for DILUTION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.
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The standard EQUATION for DILUTION compares the INITIAL and FINAL CONDITIONS of the SOLUTION, as shown by the expression: (π1 )(π1 ) = (π2 )(π2 ) Where: β’ π is the MOLARITY or CONCENTRATION given in units of moles per liter (πππ/πΏ) β’ π is the VOLUME given in units of liters (πΏ) β’ The subscript 1 refers to BEFORE DILUTION β’ The subscript 2 refers to AFTER DILUTION
We are told that the INITIAL VOLUME of the SOLUTION is π1 = 525 ππΏ and the INITIAL MOLARITY of the SOLUTION is π1 = 6.6 π. For the FINAL CONDITION, we are told the FINAL CONCENTRATION of the SOLUTION is π2 = 0.075 π, and need to SOLVE for the FINAL VOLUME of the SOLUTION. Plugging in the INITIAL VOLUME, INITIAL CONCENTRATION, and FINAL CONCENTRATION into the FORMULA for DILUTION, we can re-write the EQUATION for DILUTION as: (5.5 π)(525 ππΏ) = (0.075 π)(π2 )
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SOLVING for the FINAL VOLUME of the SOLUTION, we find:
π2 =
(5.5 π)(525 ππΏ) = 38,500 ππΏ = 38.5 πΏ (0.075 π)
Therefore, the correct answer choice is B. ππ. π
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PROBLEM 6: The amount of water that would need to be added to 250 mL of a 1.6 M HCl solution to make a 0.55 M solution is most close to: A. 315 B. 328 C. 419 D. 477
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SOLUTION 6: The TOPIC of DILUTION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. In this problem, we are GIVEN the INITIAL CONCENTRATION and VOLUME of a SOLUTION, and then asked to SOLVE for the AMOUNT of WATER that would need to be ADDED to CHANGE the MOLARITY to the DEFINED value in the PROBLEM STATEMENT. When working with SOLUTIONS in CHEMISTRY, it is common to need to CHANGE the CONCENTRATION of a SOLUTION by changing the AMOUNT of SOLVE. A DILUTION is the ADDITION of SOLVENT to a SOLUTION, which DECREASES the CONCENTRATION of the SOLUTE in the SOLUTION. DILUTION alters the MOLARITY or CONCENTRATION of the SOLUTION such that the NUMBER of MOLES of SOLUTE can be expressed as: πππππ ππ ππππ’π‘π = (πππππππ‘π¦)(πΏππ‘πππ ππ ππππ’π‘πππ) = (π)(π) The FORMULA for DILUTION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. Made with
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The standard EQUATION for DILUTION compares the INITIAL and FINAL CONDITIONS of the SOLUTION, as shown by the expression: (π1 )(π1 ) = (π2 )(π2 ) Where: β’ π is the MOLARITY or CONCENTRATION given in units of moles per liter (πππ/πΏ) β’ π is the VOLUME given in units of liters (πΏ) β’ The subscript 1 refers to BEFORE DILUTION β’ The subscript 2 refers to AFTER DILUTION
We are told that the INITIAL VOLUME of the SOLUTION is π1 = 250 ππΏ and the INITIAL MOLARITY of the SOLUTION is π1 = 1.6 π. For the FINAL CONDITION, we are told the FINAL CONCENTRATION of the SOLUTION is π2 = 0.55 π, and need to SOLVE for the FINAL VOLUME of the SOLUTION. Plugging in the INITIAL VOLUME, INITIAL CONCENTRATION, and FINAL CONCENTRATION into the FORMULA for DILUTION, we can re-write the EQUATION for DILUTION as: (1.6 π)(250 ππΏ) = (0.55 π)(π2 )
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SOLVING for the FINAL VOLUME of the SOLUTION, we find:
π2 =
(1.6 π)(250 ππΏ) = 727.27 ππΏ (0.55 π)
As the PROBLEM STATEMENT is asking for how much WATER should be ADDED, we simply CALCULATE the DIFFERENCE between the INITIAL VOLUME and the FINAL VOLUME: π₯ππππ’ππ = π2 β π1 = 727.27 ππΏ β 250 ππΏ = 477.27 ππΏ
Therefore, the correct answer choice is D. πππ
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PROBLEM 1: Environmental engineers typically do calculations involving the mixture of water samples and various chemical additives to measure certain physical characteristics. Given water is added to 25 mL of a 1.25 M πΎ2 ππ4 stock solution, the amount of water required to result in a 500 mL solution having a potassium ion concentration of 0.500 M is most close to: A. 250 ππΏ B. 350 ππΏ C. 375 ππΏ D. 475 ππΏ
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SOLUTION 1: The TOPIC of SOLUTION CONCENTRATIONS is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.
In this problem, we are looking to SOLVE for the AMOUNT of WATER that should be ADDED to in a CONCENTRATION of POTASSIUM IONS (πΎ + ) of 0.500 M. As we are GIVEN the CONCENTRATIONS and VOLUME of each SUBSTANCE, we can USE a STOICHIOMETRIC RATIO to determine the CONCENTRATION of POTASSIUM IONS (πΎ + ). However, before can SOLVE for the CONCENTRATION of POTASSIUM IONS (πΎ + ), we need to CALCULATE the CONCENTRATION of the STOCK SOLUTION in the overall SOLUTION. As we are working with STOCK SOLUTION of πΎ2 ππ4, we know that the MAJORITY COMPONENT of the SOLUTION will be the WATER that is MIXED in to DILUTE the STOCK SOLUTION. Therefore, the STOCK SOLUTION of πΎ2 ππ4 is the SOLUTE, and the WATER being ADDED is the SOLVENT.
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The term CONCENTRATION is used to INDICATE the AMOUNT of SOLUTE dissolved in a given quantity of SOLVENT or SOLUTION. Brackets β[ ]β are commonly used to indicate that a value REPRESENTS the CONCENTRATION of a SUBSTANCE, and have associated UNITS of MOLES of SOLUTE per LITER of SOLUTION. The CONCENTRATION is defined as the PERCENTAGE of the SOLUTE present within a given VOLUME of SOLUTION.
πΆππππππ‘πππ‘πππ =
ππππ ππ ππππ’π‘π πΏππ‘ππ ππ ππππ’π‘πππ
The CONCENTRATION represents the PHYSICAL PARAMETER of VOLUME and is commonly expressed as the MOLARITY or MOLALITY of a SOLUTION. When the CONCENTRATION is expressed on a VOLUMETRIC SCALE such that the AMOUNT of SOLUTE contained in a given VOLUME of SOLUTION is EQUAL to the PRODUCT of the VOLUME, the CONCENTRATION is given as: π΄πππ’ππ‘ ππ ππππ’π‘π = ππππ’ππ Γ πΆππππππ‘πππ‘πππ We should NOTE that the VOLUME is given in UNITS of MILLILITERS, so we need to take note of making sure all of our UNITS are CONSISTENT.
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The TABLE of METRIC PREFIXES can be referenced under the SUBJECT of UNITS on page 1 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Using the METRIX PREFIX for milli-, the CONVERSION FACTOR for converting LITERS (L) to MILLILITERS (mL) is given as: 1 ππΏ = 1 Γ 10β3 πΏ As we are told that 25 mL of a 1.25 M πΎ2 ππ4 stock solution is being used, we can SOLVE the AMOUNT of SOLUTE in the SOLUTION as: 1 Γ 10β3 πΏ 1.25 πππ πΎ2 ππ4 π΄πππ’ππ‘ ππ ππππ’π‘π = 25 ππΏ Γ Γ = 0.03125 πππ πΎ2 ππ4 1 ππΏ 1πΏ Now that we know the AMOUNT of STOCK SOLUTION we are working with, we need to take it one step further to CALCULATED the amount of POTASSIUM IONS (πΎ + ) in the SOLUTION. We can then use a MOLAR RATIO to COMPARE the number of MOLES of POTASSIUM (πΎ + ) in a MOLECULE of πΎ2 ππ4 . As there are 2 POTASSIUMS IONS (πΎ + ) in every MOLECULE of πΎ2 ππ4, we can CALCULATE the CONCENTRATIONS of POTASSIUMS IONS [πΎ + ] in the SOLUTE as:
[πΎ + ]
2 ππππ πΎ + = 0.03125 ππΏ πΎ2 ππ4 Γ = 0.0625 πππ πΎ + 1 πππ πΎ2 ππ4 Made with
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Recall the CONCENTRATION is defined as the PERCENTAGE of the SOLUTE present within a given VOLUME of SOLUTION. Based on the DEFINED concentration of [πΎ + ] = 0.5 π from the PROBLEM STATEMENT, we can RELATE the NUMBER of MOLES of POTASSIUM IONS to an VOLUME representing the MOLES of SOLUTE:
0.5 π
[πΎ + ]
ππππ πΎ + ππ π πππ’π‘π = 1 πΏ ππ π πππ£πππ‘
Now that we have defined CONCENTRATION of POTASSIUM IONS in the SOLUTE as [πΎ + ] = 0.0625 πππ πΎ + , we can PLUG in the GIVEN potassium ion concentration level of 0.5 M to calculate the MOLS of SOLUTE in TERMS of VOLUME as:
0.0625 πππ πΎ + Γ
1 πΏ ππ π πππ£πππ‘ = 0.125 πΏ = 125 ππΏ 0.500 πππ πΎ +
As we are told the VOLUME of the SOLUTION is 500 mL, we can subtract the VOLUME of the POTASSIUM IONS to calculate the VOLUME of water ADDED as: 500 ππΏ β 125 ππ = 375 ππΏ
Therefore, the correct answer choice is C. 375 mL
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PROBLEM 2: Given a sample solution with a volume of 375 mL that is measured to contain 0.278 moles of solute, the molarity of the solution is most close to: A. 0.218 B. 0.378 C. 0.741 D. 0.812
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SOLUTION 2: The TOPIC of MOLARITY OF SOLUTIONS can be referenced under SUBJECT of CHEMISTRY on page 54 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. In this problem, we are looking to SOLVE for the MOLARITY of a SOLUTION given the MOLES of SOLUTE and VOLUME of SOLUTION. The MOLARITY (M) of a SOLUTION is the NUMBER of GRAM MOLES of a substance DISSOLVED in a LITER VOLUME of SOLUTION.
πππππππ‘π¦ (π) =
ππππ ππ ππππ’π‘π πΏππ‘ππ ππ ππππ’π‘πππ
We are TOLD that the VOLUME of the SOLUTION sample is 375 mL. As the UNITS of MOLARITY are in LITERS (L), we need to CONVERT the VOLUME from milliliters (mL) to liters (L). Doing so, we find the VOLUME in UNITS of milliliters (mL) is: 1 Γ 10β3 πΏ ππππ’ππ = 375 ππΏ Γ = 0.375 πΏ 1 ππΏ We are also TOLD that the MOLES of SOLUTE in the SOLUTION is 0.278 moles, which we can PLUG directly into the FORMULA for MOLARITY. Made with
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As we now have the VOLUME in liters, and are given the MOLES of SOLUTE, we can simply PLUG and CHUG these VALUES in the FORMULA for MOLARITY and find:
πππππππ‘π¦ (π) =
ππππ ππ ππππ’π‘π 0.278 ππππ πππ = = 0.741 = 0.741 π πΏππ‘ππ ππ ππππ’π‘πππ 0.375 πΏ πΏ
Therefore, the correct answer choice is C. 0.741
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PROBLEM 3: Given 14.38 moles of potassium fluoride is dissolved to make 13 mL of solution, the molarity of the solution is most close to: A. 854 B. 1106 C. 1352 D. 2357
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SOLUTION 3: The TOPIC of MOLARITY OF SOLUTIONS can be referenced under SUBJECT of CHEMISTRY on page 54 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. In this problem, we are looking to SOLVE for the MOLARITY of a SOLUTION given the MOLES of SOLUTE and VOLUME of SOLUTION. The MOLARITY (M) of a SOLUTION is the NUMBER of GRAM MOLES of a substance DISSOLVED in a LITER VOLUME of SOLUTION.
πππππππ‘π¦ (π) =
ππππ ππ ππππ’π‘π πΏππ‘ππ ππ ππππ’π‘πππ
We are TOLD that the VOLUME of the SOLUTION sample is 13 mL. As the UNITS of MOLARITY are in LITERS (L), we need to CONVERT the VOLUME from milliliters (mL) to liters (L). Doing so, we find the VOLUME in UNITS of milliliters (mL) is: 1 Γ 10β3 πΏ ππππ’ππ = 13 ππΏ Γ = 0.013 πΏ 1 ππΏ We are also TOLD that the MOLES of SOLUTE in the SOLUTION is 14.38 moles, which we can PLUG directly into the FORMULA for MOLARITY. Made with
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As we now have the VOLUME in liters, and are given the MOLES of SOLUTE, we can simply PLUG and CHUG these VALUES in the FORMULA for MOLARITY and find:
πππππππ‘π¦ (π) =
ππππ ππ ππππ’π‘π 14.38 ππππ πΎπΉ πππ = = 1106.15 = 1106 π πΏππ‘ππ ππ ππππ’π‘πππ 0.013 πΏ πΏ
Therefore, the correct answer choice is B. 1106
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PROBLEM 4: If 25 mL of water is added to 375 mL of a 0.65 M peroxide solution, the molarity of the diluted solution is most close to: A. 0.12 B. 0.28 C. 0.53 D. 0.61
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SOLUTION 4: The TOPIC of DILUTION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. In this problem, we are GIVEN the INITIAL CONCENTRATION and VOLUME of a SOLUTION, and then asked to SOLVE for the FINAL CONCENTRATION of the SOLUTION, once some DEFINED AMOUNT of SOLVENT is ADDED to the SOLUTION. When working with SOLUTIONS in CHEMISTRY, it is common to need to CHANGE the CONCENTRATION of a SOLUTION by changing the AMOUNT of SOLVE. A DILUTION is the ADDITION of SOLVENT to a SOLUTION, which DECREASES the CONCENTRATION of the SOLUTE in the SOLUTION. DILUTION alters the MOLARITY or CONCENTRATION of the SOLUTION such that the NUMBER of MOLES of SOLUTE can be expressed as: πππππ ππ ππππ’π‘π = (πππππππ‘π¦)(πΏππ‘πππ ππ ππππ’π‘πππ) = (π)(π) The FORMULA for DILUTION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. Made with
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The standard EQUATION for DILUTION compares the INITIAL and FINAL CONDITIONS of the SOLUTION, as shown by the expression: (π1 )(π1 ) = (π2 )(π2 ) Where: β’ π is the MOLARITY or CONCENTRATION given in units of moles per liter (πππ/πΏ) β’ π is the VOLUME given in units of liters (πΏ) β’ The subscript 1 refers to BEFORE DILUTION β’ The subscript 2 refers to AFTER DILUTION
We are told that the INITIAL VOLUME of the SOLUTION is π1 = 375 ππΏ and the INITIAL MOLARITY of the SOLUTION is π1 = 0.65 π. For the FINAL CONDITION, we are told that the VOLUME is to INCREASE by 25 mL, and then asked to SOLVE for the FINAL CONCENTRATION of the SOLUTION. We can CALCULATE the FINAL VOLUME (π2 ), as the SUM of the INITIAL VOLUME plus the ADDITION of 25 mL of SOLVENT. π2 = 375 ππΏ + 25 ππΏ = 400 ππΏ
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Plugging in the INITIAL VOLUME, INITIAL CONCENTRATION, and FINAL VOLUME into the FORMULA for DILUTION, we can re-write the EQUATION for DILUTION as: (0.65 π)(375 ππΏ) = π2 (400 ππΏ) SOLVING for the FINAL CONCENTRATION of the SOLUTION, we find:
π2 =
(0.65 π)(375 ππΏ) = 0.609 π 400 ππΏ
Therefore, the correct answer choice is D. 0.61
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PROBLEM 5: The amount of 0.075 M HF solution that can be made by diluting 525 mL of 5.5 M HF in units of liters is most close to: A. 37.0 B. 38.5 C. 42.5 D. 44.0
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SOLUTION 5: The TOPIC of DILUTION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. In this problem, we are GIVEN the INITIAL CONCENTRATION and VOLUME of a SOLUTION, and then asked to SOLVE for the FINAL VOLUME of the SOLUTION, GIVEN the FINAL CONCENTRATION of the SOLUTION. When working with SOLUTIONS in CHEMISTRY, it is common to need to CHANGE the CONCENTRATION of a SOLUTION by changing the AMOUNT of SOLVE. A DILUTION is the ADDITION of SOLVENT to a SOLUTION, which DECREASES the CONCENTRATION of the SOLUTE in the SOLUTION. DILUTION alters the MOLARITY or CONCENTRATION of the SOLUTION such that the NUMBER of MOLES of SOLUTE can be expressed as: πππππ ππ ππππ’π‘π = (πππππππ‘π¦)(πΏππ‘πππ ππ ππππ’π‘πππ) = (π)(π) The FORMULA for DILUTION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.
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The standard EQUATION for DILUTION compares the INITIAL and FINAL CONDITIONS of the SOLUTION, as shown by the expression: (π1 )(π1 ) = (π2 )(π2 ) Where: β’ π is the MOLARITY or CONCENTRATION given in units of moles per liter (πππ/πΏ) β’ π is the VOLUME given in units of liters (πΏ) β’ The subscript 1 refers to BEFORE DILUTION β’ The subscript 2 refers to AFTER DILUTION
We are told that the INITIAL VOLUME of the SOLUTION is π1 = 525 ππΏ and the INITIAL MOLARITY of the SOLUTION is π1 = 6.6 π. For the FINAL CONDITION, we are told the FINAL CONCENTRATION of the SOLUTION is π2 = 0.075 π, and need to SOLVE for the FINAL VOLUME of the SOLUTION. Plugging in the INITIAL VOLUME, INITIAL CONCENTRATION, and FINAL CONCENTRATION into the FORMULA for DILUTION, we can re-write the EQUATION for DILUTION as: (5.5 π)(525 ππΏ) = (0.075 π)(π2 )
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SOLVING for the FINAL VOLUME of the SOLUTION, we find:
π2 =
(5.5 π)(525 ππΏ) = 38,500 ππΏ = 38.5 πΏ (0.075 π)
Therefore, the correct answer choice is B. ππ. π
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PROBLEM 6: The amount of water that would need to be added to 250 mL of a 1.6 M HCl solution to make a 0.55 M solution is most close to: A. 315 B. 328 C. 419 D. 477
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SOLUTION 6: The TOPIC of DILUTION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. In this problem, we are GIVEN the INITIAL CONCENTRATION and VOLUME of a SOLUTION, and then asked to SOLVE for the AMOUNT of WATER that would need to be ADDED to CHANGE the MOLARITY to the DEFINED value in the PROBLEM STATEMENT. When working with SOLUTIONS in CHEMISTRY, it is common to need to CHANGE the CONCENTRATION of a SOLUTION by changing the AMOUNT of SOLVE. A DILUTION is the ADDITION of SOLVENT to a SOLUTION, which DECREASES the CONCENTRATION of the SOLUTE in the SOLUTION. DILUTION alters the MOLARITY or CONCENTRATION of the SOLUTION such that the NUMBER of MOLES of SOLUTE can be expressed as: πππππ ππ ππππ’π‘π = (πππππππ‘π¦)(πΏππ‘πππ ππ ππππ’π‘πππ) = (π)(π) The FORMULA for DILUTION is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. Made with
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The standard EQUATION for DILUTION compares the INITIAL and FINAL CONDITIONS of the SOLUTION, as shown by the expression: (π1 )(π1 ) = (π2 )(π2 ) Where: β’ π is the MOLARITY or CONCENTRATION given in units of moles per liter (πππ/πΏ) β’ π is the VOLUME given in units of liters (πΏ) β’ The subscript 1 refers to BEFORE DILUTION β’ The subscript 2 refers to AFTER DILUTION
We are told that the INITIAL VOLUME of the SOLUTION is π1 = 250 ππΏ and the INITIAL MOLARITY of the SOLUTION is π1 = 1.6 π. For the FINAL CONDITION, we are told the FINAL CONCENTRATION of the SOLUTION is π2 = 0.55 π, and need to SOLVE for the FINAL VOLUME of the SOLUTION. Plugging in the INITIAL VOLUME, INITIAL CONCENTRATION, and FINAL CONCENTRATION into the FORMULA for DILUTION, we can re-write the EQUATION for DILUTION as: (1.6 π)(250 ππΏ) = (0.55 π)(π2 )
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SOLVING for the FINAL VOLUME of the SOLUTION, we find:
π2 =
(1.6 π)(250 ππΏ) = 727.27 ππΏ (0.55 π)
As the PROBLEM STATEMENT is asking for how much WATER should be ADDED, we simply CALCULATE the DIFFERENCE between the INITIAL VOLUME and the FINAL VOLUME: π₯ππππ’ππ = π2 β π1 = 727.27 ππΏ β 250 ππΏ = 477.27 ππΏ
Therefore, the correct answer choice is D. πππ
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