D) If you have 100.0 mL of solution, how many moles of NaCl are present? Wha the mass? (MM of NaCl is 58.44 g/mol) mol M= mol=(0@5OM) CO.100OL) = 0.0050 ao( 5%.4)-0.2a mol= MVc 100.mL=01000L 2 mol E) You have diluted 5.00 mL of the original solution to 25.00 mL. What is the concentration now?

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### Problem D:
If you have 100.0 mL of solution, how many moles of NaCl are present? What is the mass? (Molar Mass of NaCl is 58.44 g/mol)

#### Solution:

1. Formula: \( M = \frac{\text{mol}}{V} \)
2. Given: \( V = 100.0 \, \text{mL} = 0.1000 \, \text{L} \)
3. Moles Calculation: 
   \[
   \text{mol} = (0.050 \,M)(0.1000 \,L) = 0.0050 \, \text{mol}
   \]
4. Mass Calculation:
   \[
   = 0.0050 \, \text{mol} \times \frac{58.44 \,\text{g}}{1 \,\text{mol}} = 0.292 \,\text{g NaCl}
   \]

### Problem E:
You have diluted 5.00 mL of the original solution to 25.00 mL. What is the concentration now?

#### Solution:

1. Initial Conditions:
   - \( V_1 = 5.00 \, \text{mL} \)
   - \( V_2 = 25.00 \, \text{mL} \)
   - \( M_1 = 0.050 \, M \)
   - \( M_2 = ? \)

2. Using the dilution formula: \( V_1 M_1 = V_2 M_2 \)
   \[
   (5.00)(0.050) = (25.00)(M_2)
   \]
   \[
   M_2 = \frac{(5.00)(0.050)}{25} = 0.010 \, M
   \]

### Explanation:
- **Graph/Diagram**: There are no graphs or diagrams in the given text, only calculation steps and formulas written by hand.
- **Methods Utilized**: The calculations use standard formulas for molarity and dilution, illustrating the step-by-step process to find the number of moles and the diluted concentration.
Transcribed Image Text:### Problem D: If you have 100.0 mL of solution, how many moles of NaCl are present? What is the mass? (Molar Mass of NaCl is 58.44 g/mol) #### Solution: 1. Formula: \( M = \frac{\text{mol}}{V} \) 2. Given: \( V = 100.0 \, \text{mL} = 0.1000 \, \text{L} \) 3. Moles Calculation: \[ \text{mol} = (0.050 \,M)(0.1000 \,L) = 0.0050 \, \text{mol} \] 4. Mass Calculation: \[ = 0.0050 \, \text{mol} \times \frac{58.44 \,\text{g}}{1 \,\text{mol}} = 0.292 \,\text{g NaCl} \] ### Problem E: You have diluted 5.00 mL of the original solution to 25.00 mL. What is the concentration now? #### Solution: 1. Initial Conditions: - \( V_1 = 5.00 \, \text{mL} \) - \( V_2 = 25.00 \, \text{mL} \) - \( M_1 = 0.050 \, M \) - \( M_2 = ? \) 2. Using the dilution formula: \( V_1 M_1 = V_2 M_2 \) \[ (5.00)(0.050) = (25.00)(M_2) \] \[ M_2 = \frac{(5.00)(0.050)}{25} = 0.010 \, M \] ### Explanation: - **Graph/Diagram**: There are no graphs or diagrams in the given text, only calculation steps and formulas written by hand. - **Methods Utilized**: The calculations use standard formulas for molarity and dilution, illustrating the step-by-step process to find the number of moles and the diluted concentration.
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