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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### Calculating the Volume of a Silver Nitrate Solution
**Problem Statement:**
If 20. g of \( \text{AgNO}_3 \) is available, what volume of 0.20 M \( \text{AgNO}_3 \) solution can be prepared?
**Solution:**
---
#### Given:
- **Mass of \( \text{AgNO}_3 \)** = 20 g
- **Molarity (M)** of the solution = 0.20 M
#### Required:
- **Volume of the solution** in mL
**Formula to use:**
\[ \text{Volume (L)} = \frac{\text{Mass of } \text{AgNO}_3 \text{ (g)}}{\text{Molarity (M)} \times \text{Molar Mass of } \text{AgNO}_3 \text{ (g/mol)}} \]
**Molar Mass of \( \text{AgNO}_3 \):**
- Silver (Ag): 107.87 g/mol
- Nitrogen (N): 14.01 g/mol
- Oxygen (O): 16.00 g/mol (since there are 3 oxygen atoms, it's 3 x 16.00)
So,
\[ \text{Molar Mass of } \text{AgNO}_3 = 107.87 + 14.01 + (3 \times 16) \ g/mol \]
**Calculate the molar mass:**
\[ \text{Molar Mass of } \text{AgNO}_3 = 107.87 + 14.01 + 48.00 \ g/mol \]
\[ \text{Molar Mass of } \text{AgNO}_3 = 169.88 \ g/mol \]
**Calculate the volume:**
\[ \text{Volume (L)} = \frac{20 \ \text{g}}{0.20 \ \text{M} \times 169.88 \ \text{g/mol}} \]
\[ \text{Volume (L)} = \frac{20}{33.976} \]
\[ \text{Volume (L)} \approx 0.588 \]
To convert this volume to mL:
\[ \text{Volume (mL)} = 0.588 \ \text{L} \times 1000 \ \text{mL/L} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3defd924-4235-4c0f-8415-f3b777029d52%2F1fb643e9-1fd2-4d8d-8cd4-8b601a6b12ea%2Fxsi7eqk_processed.png&w=3840&q=75)
Transcribed Image Text:---
### Calculating the Volume of a Silver Nitrate Solution
**Problem Statement:**
If 20. g of \( \text{AgNO}_3 \) is available, what volume of 0.20 M \( \text{AgNO}_3 \) solution can be prepared?
**Solution:**
---
#### Given:
- **Mass of \( \text{AgNO}_3 \)** = 20 g
- **Molarity (M)** of the solution = 0.20 M
#### Required:
- **Volume of the solution** in mL
**Formula to use:**
\[ \text{Volume (L)} = \frac{\text{Mass of } \text{AgNO}_3 \text{ (g)}}{\text{Molarity (M)} \times \text{Molar Mass of } \text{AgNO}_3 \text{ (g/mol)}} \]
**Molar Mass of \( \text{AgNO}_3 \):**
- Silver (Ag): 107.87 g/mol
- Nitrogen (N): 14.01 g/mol
- Oxygen (O): 16.00 g/mol (since there are 3 oxygen atoms, it's 3 x 16.00)
So,
\[ \text{Molar Mass of } \text{AgNO}_3 = 107.87 + 14.01 + (3 \times 16) \ g/mol \]
**Calculate the molar mass:**
\[ \text{Molar Mass of } \text{AgNO}_3 = 107.87 + 14.01 + 48.00 \ g/mol \]
\[ \text{Molar Mass of } \text{AgNO}_3 = 169.88 \ g/mol \]
**Calculate the volume:**
\[ \text{Volume (L)} = \frac{20 \ \text{g}}{0.20 \ \text{M} \times 169.88 \ \text{g/mol}} \]
\[ \text{Volume (L)} = \frac{20}{33.976} \]
\[ \text{Volume (L)} \approx 0.588 \]
To convert this volume to mL:
\[ \text{Volume (mL)} = 0.588 \ \text{L} \times 1000 \ \text{mL/L} \]
![**Question:**
A solution was prepared by mixing \( 40.00 \) mL of \( 0.100 \, M \) \( \text{HNO}_3 \) and \( 170.00 \) mL of \( 0.200 \, M \) \( \text{HNO}_3 \). Calculate the molarity of the final solution of nitric acid.
Molarity = ____ \( M \)
**Explanation:**
To calculate the final molarity of the nitric acid solution, you need to use the formula for combined solutions:
\[ C_{\text{final}} = \frac{(C_1 \times V_1) + (C_2 \times V_2)}{V_{\text{total}}} \]
Where:
- \( C_1 \) and \( C_2 \) are the initial concentrations of the two solutions.
- \( V_1 \) and \( V_2 \) are the volumes of the two solutions.
- \( V_{\text{total}} \) is the total volume of the final solution.
In this example:
- \( C_1 = 0.100 \, M \)
- \( V_1 = 40.00 \, mL \)
- \( C_2 = 0.200 \, M \)
- \( V_2 = 170.00 \, mL \)
Calculate the moles of \( \text{HNO}_3 \) in each solution:
- Moles in first solution \( = C_1 \times V_1 = 0.100 \, M \times 40.00 \times 10^{-3} \, L = 0.00400 \, \text{moles} \)
- Moles in second solution \( = C_2 \times V_2 = 0.200 \, M \times 170.00 \times 10^{-3} \, L = 0.03400 \, \text{moles} \)
Total moles of \( \text{HNO}_3 = 0.00400 + 0.03400 = 0.03800 \, \text{moles} \)
Total volume \( V_{\text{total}} = V_1 + V_2 = 40.00 \, mL + 170.00 \, mL =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3defd924-4235-4c0f-8415-f3b777029d52%2F1fb643e9-1fd2-4d8d-8cd4-8b601a6b12ea%2Fwwy3l9v_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
A solution was prepared by mixing \( 40.00 \) mL of \( 0.100 \, M \) \( \text{HNO}_3 \) and \( 170.00 \) mL of \( 0.200 \, M \) \( \text{HNO}_3 \). Calculate the molarity of the final solution of nitric acid.
Molarity = ____ \( M \)
**Explanation:**
To calculate the final molarity of the nitric acid solution, you need to use the formula for combined solutions:
\[ C_{\text{final}} = \frac{(C_1 \times V_1) + (C_2 \times V_2)}{V_{\text{total}}} \]
Where:
- \( C_1 \) and \( C_2 \) are the initial concentrations of the two solutions.
- \( V_1 \) and \( V_2 \) are the volumes of the two solutions.
- \( V_{\text{total}} \) is the total volume of the final solution.
In this example:
- \( C_1 = 0.100 \, M \)
- \( V_1 = 40.00 \, mL \)
- \( C_2 = 0.200 \, M \)
- \( V_2 = 170.00 \, mL \)
Calculate the moles of \( \text{HNO}_3 \) in each solution:
- Moles in first solution \( = C_1 \times V_1 = 0.100 \, M \times 40.00 \times 10^{-3} \, L = 0.00400 \, \text{moles} \)
- Moles in second solution \( = C_2 \times V_2 = 0.200 \, M \times 170.00 \times 10^{-3} \, L = 0.03400 \, \text{moles} \)
Total moles of \( \text{HNO}_3 = 0.00400 + 0.03400 = 0.03800 \, \text{moles} \)
Total volume \( V_{\text{total}} = V_1 + V_2 = 40.00 \, mL + 170.00 \, mL =
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