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The problem asks for the volume in liters of a 0.173 M manganese(II) sulfate solution that contains 71.2 g of solute. Here's how to solve it step by step:

### Steps:
1. **Molar Mass of Manganese(II) Sulfate**:
   First, find the molar mass of manganese(II) sulfate, \( \text{MnSO}_4 \). 
   - Manganese (Mn): 54.94 g/mol
   - Sulfur (S): 32.06 g/mol
   - Oxygen (O): \( 16.00 \times 4 = 64.00 \) g/mol

   So, the molar mass of \( \text{MnSO}_4 = 54.94 + 32.06 + 64.00 = 151.00 \) g/mol.

2. **Moles of Manganese(II) Sulfate**:
   Now, calculate the number of moles of manganese(II) sulfate in 71.2 g of solute:
   \[
   \text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{71.2 \, \text{g}}{151.00 \, \text{g/mol}} = 0.471 \, \text{mol}
   \]

3. **Volume of Solution**:
   Use the molarity formula to find the volume:
   \[
   \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}
   \]
   Rearranging for volume:
   \[
   \text{volume} = \frac{\text{moles of solute}}{\text{Molarity}} = \frac{0.471 \, \text{mol}}{0.173 \, \text{M}} = 2.72 \, \text{L}
   \]

Thus, the volume of the solution is approximately **2.72 liters**.

Would you like more details or have any questions?

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Here are 5 related questions you might find helpful:
1. How do you calculate molarity from a given volume and amount of solute?
2. What is the importance of molar mass in solution calculations?
3. How would the volume change if the molarity were different?
4. How do significant figures affect the final answer in molarity problems?
5. How does molarity relate to other concentration units like percent concentration?

**Tip:** Always check the units to ensure consistency (grams to moles, liters, etc.) when calculating molarity and volume.

Calculate the volume in liters of 0.173 M manganese(II) sulfate solution that contains 71.2 g of solute.

Learn how to calculate the volume of a 0.173 M manganese(II) sulfate solution that contains 71.2 grams of solute. This step-by-step solution covers molarity calculations, moles, and molar mass determination, perfect for high school chemistry students.

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