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Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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![### Determining the Molarity of a Manganese(II) Chloride Solution
A student weighs out a 3.48 g sample of manganese(II) chloride, transfers it to a 125 mL volumetric flask, adds enough water to dissolve it, and then adds water to the 125 mL tic mark.
**Problem:**
What is the molarity of MnCl₂ in the resulting solution?
**Solution:**
To find the molarity, we need to follow these steps:
1. **Calculate the molar mass of MnCl₂:**
- Manganese (Mn): Atomic mass = 54.94 g/mol
- Chlorine (Cl): Atomic mass = 35.45 g/mol (since there are two Cl atoms, multiply by 2)
Molar mass of MnCl₂ = 54.94 g/mol + (2 × 35.45 g/mol) = 54.94 g/mol + 70.90 g/mol = 125.84 g/mol
2. **Convert grams of MnCl₂ to moles:**
\[\text{moles of MnCl₂} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{3.48 \, \text{g}}{125.84 \, \text{g/mol}} = 0.02765 \, \text{mol}\]
3. **Convert the volume from mL to L:**
\[125 \, \text{mL} = 0.125 \, \text{L}\]
4. **Calculate the molarity (M):**
\[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{0.02765 \, \text{mol}}{0.125 \, \text{L}} = 0.2212 \, \text{M}\]
Therefore, the molarity of MnCl₂ in the resulting solution is **0.2212 M**.
**Box for Answer:**
\[ \boxed{0.2212 \, \text{M}} \]
This question can be useful for students who want to understand the process of calculating the molarity of a solution from a given mass of solute and the final volume of solution.](https://content.bartleby.com/qna-images/question/a527ffdc-16df-4a1f-9953-96956a75d05e/b472def2-87c3-4b72-8ac7-efc81b79096d/o7r8ih_thumbnail.png)
Transcribed Image Text:### Determining the Molarity of a Manganese(II) Chloride Solution
A student weighs out a 3.48 g sample of manganese(II) chloride, transfers it to a 125 mL volumetric flask, adds enough water to dissolve it, and then adds water to the 125 mL tic mark.
**Problem:**
What is the molarity of MnCl₂ in the resulting solution?
**Solution:**
To find the molarity, we need to follow these steps:
1. **Calculate the molar mass of MnCl₂:**
- Manganese (Mn): Atomic mass = 54.94 g/mol
- Chlorine (Cl): Atomic mass = 35.45 g/mol (since there are two Cl atoms, multiply by 2)
Molar mass of MnCl₂ = 54.94 g/mol + (2 × 35.45 g/mol) = 54.94 g/mol + 70.90 g/mol = 125.84 g/mol
2. **Convert grams of MnCl₂ to moles:**
\[\text{moles of MnCl₂} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{3.48 \, \text{g}}{125.84 \, \text{g/mol}} = 0.02765 \, \text{mol}\]
3. **Convert the volume from mL to L:**
\[125 \, \text{mL} = 0.125 \, \text{L}\]
4. **Calculate the molarity (M):**
\[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{0.02765 \, \text{mol}}{0.125 \, \text{L}} = 0.2212 \, \text{M}\]
Therefore, the molarity of MnCl₂ in the resulting solution is **0.2212 M**.
**Box for Answer:**
\[ \boxed{0.2212 \, \text{M}} \]
This question can be useful for students who want to understand the process of calculating the molarity of a solution from a given mass of solute and the final volume of solution.
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