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. What is the molarity of MnCl, in the resulting 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. What is the molarity of MnCl, in the resulting solution?
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
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa527ffdc-16df-4a1f-9953-96956a75d05e%2Fb472def2-87c3-4b72-8ac7-efc81b79096d%2Fo7r8ih.png&w=3840&q=75)
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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