## Solubility and Concentration Calculations ### 5. Effect of Temperature on Solubility - **Solids**: Increasing the temperature of a solution generally increases the solubility of solids. - **Gases**: As the temperature of a solution increases, gases become less soluble. ### 6. Calculating Required Volume for Reactions - **Problem**: What volume of 1.50 M NaCl is needed for a reaction requiring 146.3 g of NaCl? - **Solution**: - Molar mass of NaCl: \( \text{Na} = 22.99 \, \text{g/mol}, \, \text{Cl} = 35.45 \, \text{g/mol} \) - Total: \( 22.99 + 35.45 = 58.44 \, \text{g/mol} \) - Volume required: 1.67 L ### 7. Molarity Calculation - **Problem**: What is the molarity of a solution with 8.2 g of potassium chromate (\( \text{K}_2\text{CrO}_4 \)) in 500. mL? - **Solution**: - Molar mass \( \text{K}_2\text{CrO}_4 \): - \( 2 \times 39.10 (\text{K}) + 52.00 (\text{Cr}) + 4 \times 16.00 (\text{O}) = 194.2 \, \text{g/mol} \) - Moles of \( \text{K}_2\text{CrO}_4 \): \( \frac{8.2 \, \text{g}}{194.2 \, \text{g/mol}} = 0.042 \, \text{mol} \) - Volume in liters: \( \frac{500 \, \text{mL}}{1000 \, \text{mL/L}} = 0.500 \, \text{L} \) - Molarity: \( \frac{0.042 \, \text{mol}}{0.500 \, \text{L}} = 0.084 \, \text{M} \) ### 8. Percent (w/w) Calculation
## Solubility and Concentration Calculations ### 5. Effect of Temperature on Solubility - **Solids**: Increasing the temperature of a solution generally increases the solubility of solids. - **Gases**: As the temperature of a solution increases, gases become less soluble. ### 6. Calculating Required Volume for Reactions - **Problem**: What volume of 1.50 M NaCl is needed for a reaction requiring 146.3 g of NaCl? - **Solution**: - Molar mass of NaCl: \( \text{Na} = 22.99 \, \text{g/mol}, \, \text{Cl} = 35.45 \, \text{g/mol} \) - Total: \( 22.99 + 35.45 = 58.44 \, \text{g/mol} \) - Volume required: 1.67 L ### 7. Molarity Calculation - **Problem**: What is the molarity of a solution with 8.2 g of potassium chromate (\( \text{K}_2\text{CrO}_4 \)) in 500. mL? - **Solution**: - Molar mass \( \text{K}_2\text{CrO}_4 \): - \( 2 \times 39.10 (\text{K}) + 52.00 (\text{Cr}) + 4 \times 16.00 (\text{O}) = 194.2 \, \text{g/mol} \) - Moles of \( \text{K}_2\text{CrO}_4 \): \( \frac{8.2 \, \text{g}}{194.2 \, \text{g/mol}} = 0.042 \, \text{mol} \) - Volume in liters: \( \frac{500 \, \text{mL}}{1000 \, \text{mL/L}} = 0.500 \, \text{L} \) - Molarity: \( \frac{0.042 \, \text{mol}}{0.500 \, \text{L}} = 0.084 \, \text{M} \) ### 8. Percent (w/w) Calculation
Chemistry
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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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I need help with my homework. Can you help me number 6 question, please?
Transcribed Image Text:## Solubility and Concentration Calculations
### 5. Effect of Temperature on Solubility
- **Solids**: Increasing the temperature of a solution generally increases the solubility of solids.
- **Gases**: As the temperature of a solution increases, gases become less soluble.
### 6. Calculating Required Volume for Reactions
- **Problem**: What volume of 1.50 M NaCl is needed for a reaction requiring 146.3 g of NaCl?
- **Solution**:
- Molar mass of NaCl: \( \text{Na} = 22.99 \, \text{g/mol}, \, \text{Cl} = 35.45 \, \text{g/mol} \)
- Total: \( 22.99 + 35.45 = 58.44 \, \text{g/mol} \)
- Volume required: 1.67 L
### 7. Molarity Calculation
- **Problem**: What is the molarity of a solution with 8.2 g of potassium chromate (\( \text{K}_2\text{CrO}_4 \)) in 500. mL?
- **Solution**:
- Molar mass \( \text{K}_2\text{CrO}_4 \):
- \( 2 \times 39.10 (\text{K}) + 52.00 (\text{Cr}) + 4 \times 16.00 (\text{O}) = 194.2 \, \text{g/mol} \)
- Moles of \( \text{K}_2\text{CrO}_4 \): \( \frac{8.2 \, \text{g}}{194.2 \, \text{g/mol}} = 0.042 \, \text{mol} \)
- Volume in liters: \( \frac{500 \, \text{mL}}{1000 \, \text{mL/L}} = 0.500 \, \text{L} \)
- Molarity: \( \frac{0.042 \, \text{mol}}{0.500 \, \text{L}} = 0.084 \, \text{M} \)
### 8. Percent (w/w) Calculation
Expert Solution
Step 1
6).
Given data :
Concentration of NaCl = 1.50 M = 1.50 mol.L-1
Mass of NaCl = 146.3 g
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