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?
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?
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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![### 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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4d6b0164-700f-47c3-a14d-0f2cc5fbb356%2F6c47b93d-e5d5-4f76-b721-2aa1340690e6%2Ful7far_processed.jpeg&w=3840&q=75)
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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