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2. Determine the volume of 0.10 M potassium thiocyanate needed to prepare 100.00 mL of a 0.00100 M solution of KSCN. 3. Determine the volume of 0.20 M iron (III) nitrate needed to prepare 50.00 mL of a 0.0100 M solution of Fe(NO3)3.

2. Determine the volume of 0.10 M potassium thiocyanate needed to prepare 100.00 mL of a 0.00100 M solution of KSCN. 3. Determine the volume of 0.20 M iron (III) nitrate needed to prepare 50.00 mL of a 0.0100 M solution of Fe(NO3)3.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Please solve this prelab question. Thanks
### Preparation of a 0.10 M HNO₃ Solution from Concentrated HNO₃ #### Instructions: 1. A 0.10 M HNO₃ solution will be prepared from concentrated HNO₃. a. **Determine the Concentration of Stock HNO₃:** - Using the method outlined in Experiment 5, determine the concentration of a stock bottle of concentrated nitric acid. - Given data: - Density = 1.42 g/mL - 70% by mass HNO₃ b. **Calculate the Volume of Concentrated HNO₃:** - Calculate the volume of concentrated HNO₃ needed to prepare 500 mL of a 0.10 M solution. #### Detailed Breakdown: **Part (a) - Determining Concentration:** To determine the molarity (M) of the concentrated stock HNO₃: 1. Calculate the mass of HNO₃ in 1 mL of solution: - Mass HNO₃ per mL = Density * % by mass - Mass HNO₃ per mL = 1.42 g/mL * 0.70 = 0.994 g/mL 2. Convert the mass of HNO₃ to moles: - Molar mass of HNO₃ ≈ 63.01 g/mol - Moles HNO₃ per mL = Mass / Molar mass - Moles HNO₃ per mL = 0.994 g / 63.01 g/mol ≈ 0.0158 mol/mL 3. Convert to molarity (moles per liter): - Molarity (M) = Moles per mL * 1000 (to convert mL to L) - Molarity = 0.0158 mol/mL * 1000 = 15.8 M **Part (b) - Calculating the Volume Needed:** To prepare 500 mL of a 0.10 M HNO₃ solution from the 15.8 M stock solution, use the dilution formula \(C_1V_1 = C_2V_2\): - \(C_1\) = Initial concentration (15.8 M) - \(V_1\) = Volume of stock solution needed -
Transcribed Image Text:### Preparation of a 0.10 M HNO₃ Solution from Concentrated HNO₃ #### Instructions: 1. A 0.10 M HNO₃ solution will be prepared from concentrated HNO₃. a. **Determine the Concentration of Stock HNO₃:** - Using the method outlined in Experiment 5, determine the concentration of a stock bottle of concentrated nitric acid. - Given data: - Density = 1.42 g/mL - 70% by mass HNO₃ b. **Calculate the Volume of Concentrated HNO₃:** - Calculate the volume of concentrated HNO₃ needed to prepare 500 mL of a 0.10 M solution. #### Detailed Breakdown: **Part (a) - Determining Concentration:** To determine the molarity (M) of the concentrated stock HNO₃: 1. Calculate the mass of HNO₃ in 1 mL of solution: - Mass HNO₃ per mL = Density * % by mass - Mass HNO₃ per mL = 1.42 g/mL * 0.70 = 0.994 g/mL 2. Convert the mass of HNO₃ to moles: - Molar mass of HNO₃ ≈ 63.01 g/mol - Moles HNO₃ per mL = Mass / Molar mass - Moles HNO₃ per mL = 0.994 g / 63.01 g/mol ≈ 0.0158 mol/mL 3. Convert to molarity (moles per liter): - Molarity (M) = Moles per mL * 1000 (to convert mL to L) - Molarity = 0.0158 mol/mL * 1000 = 15.8 M **Part (b) - Calculating the Volume Needed:** To prepare 500 mL of a 0.10 M HNO₃ solution from the 15.8 M stock solution, use the dilution formula \(C_1V_1 = C_2V_2\): - \(C_1\) = Initial concentration (15.8 M) - \(V_1\) = Volume of stock solution needed -
### Exercise Questions: **2.** Determine the volume of 0.10 M potassium thiocyanate needed to prepare 100.00 mL of a 0.00100 M solution of KSCN. **3.** Determine the volume of 0.20 M iron (III) nitrate needed to prepare 50.00 mL of a 0.0100 M solution of Fe(NO₃)₃. --- These exercises require the application of the dilution formula: \[ M_1 V_1 = M_2 V_2 \] Where: - \( M_1 \) is the molarity of the initial concentrated solution. - \( V_1 \) is the volume of the concentrated solution required. - \( M_2 \) is the molarity of the final diluted solution. - \( V_2 \) is the volume of the final diluted solution. For each exercise: 1. **Exercise 2:** - Initial molarity \( (M_1) \): 0.10 M - Final molarity \( (M_2) \): 0.00100 M - Final volume \( (V_2) \): 100.00 mL 2. **Exercise 3:** - Initial molarity \( (M_1) \): 0.20 M - Final molarity \( (M_2) \): 0.0100 M - Final volume \( (V_2) \): 50.00 mL By rearranging the dilution equation to solve for \( V_1 \) (the volume of the concentrated solution needed), the following calculations can be applied: \[ V_1 = \frac{M_2 \times V_2}{M_1} \] Substitute the given values for each problem to find \( V_1 \).
Transcribed Image Text:### Exercise Questions: **2.** Determine the volume of 0.10 M potassium thiocyanate needed to prepare 100.00 mL of a 0.00100 M solution of KSCN. **3.** Determine the volume of 0.20 M iron (III) nitrate needed to prepare 50.00 mL of a 0.0100 M solution of Fe(NO₃)₃. --- These exercises require the application of the dilution formula: \[ M_1 V_1 = M_2 V_2 \] Where: - \( M_1 \) is the molarity of the initial concentrated solution. - \( V_1 \) is the volume of the concentrated solution required. - \( M_2 \) is the molarity of the final diluted solution. - \( V_2 \) is the volume of the final diluted solution. For each exercise: 1. **Exercise 2:** - Initial molarity \( (M_1) \): 0.10 M - Final molarity \( (M_2) \): 0.00100 M - Final volume \( (V_2) \): 100.00 mL 2. **Exercise 3:** - Initial molarity \( (M_1) \): 0.20 M - Final molarity \( (M_2) \): 0.0100 M - Final volume \( (V_2) \): 50.00 mL By rearranging the dilution equation to solve for \( V_1 \) (the volume of the concentrated solution needed), the following calculations can be applied: \[ V_1 = \frac{M_2 \times V_2}{M_1} \] Substitute the given values for each problem to find \( V_1 \).
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