A perchloric acid solution is 35.00 percent perchloric acid by mass. It is simultaneously 4.4311 M at 25°C.
Compute the density of this solution in g cm-3

What volume of this solution should be used in mixing 1.00 L of a 0.63 M perchloric acid solution?
mL

### Part 1: Compute the Density of the Solution

Given:
- The solution is 35.00% perchloric acid by mass.
- Molarity (\( M \)) = 4.4311 M (moles per liter) at 25°C.

We need to calculate the density of the solution in g/cm³.

1. **Calculate the Molar Mass of Perchloric Acid (\(HClO_4\))**:
   - H: 1.008 g/mol
   - Cl: 35.45 g/mol
   - O: 16.00 g/mol (4 oxygen atoms)
   
   Molar mass of \(HClO_4\) = \( 1.008 + 35.45 + 4(16.00) = 100.458 \, \text{g/mol} \).

2. **Relate Molarity, Mass Percent, and Density**:
   - Molarity \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \)
   - Let the mass of solution be \( m_{\text{sol}} = 100 \, \text{g} \) (assuming 100 g of solution for simplicity, since percentages are easier to work with).

   From the mass percentage, the mass of perchloric acid is 35.00% of 100 g, i.e., \( m_{\text{HClO}_4} = 35.00 \, \text{g} \).

3. **Convert Mass of Perchloric Acid to Moles**:
   Using the molar mass of \(HClO_4\):
   \[
   \text{Moles of } HClO_4 = \frac{35.00 \, \text{g}}{100.458 \, \text{g/mol}} = 0.3485 \, \text{mol}.
   \]

4. **Calculate the Volume of Solution**:
   Using the molarity (4.4311 M):
   \[
   \text{Volume of solution} = \frac{\text{moles of } HClO_4}{\text{Molarity}} = \frac{0.3485 \, \text{mol}}{4.4311 \, \text{mol/L}} = 0.07866 \, \text{L} = 78.66 \, \text{mL}.
   \]

5. **Calculate the Density**:
   Density \( \rho \) is mass divided by volume:
   \[
   \rho = \frac{100 \, \text{g}}{78.66 \, \text{mL}} = 1.271 \, \text{g/mL} = 1.271 \, \text{g/cm}^3.
   \]

Thus, the density of the solution is \( \boxed{1.271 \, \text{g/cm}^3} \).

---

### Part 2: Volume of Solution to Use for Dilution

Given:
- Initial concentration \(C_1 = 4.4311 \, \text{M}\)
- Final concentration \(C_2 = 0.63 \, \text{M}\)
- Final volume \(V_2 = 1.00 \, \text{L} = 1000 \, \text{mL}\)

Using the dilution formula:
\[
C_1 V_1 = C_2 V_2
\]
\[
V_1 = \frac{C_2 V_2}{C_1} = \frac{0.63 \times 1000}{4.4311} = 142.2 \, \text{mL}
\]

Thus, the volume of the original solution needed is \( \boxed{142.2 \, \text{mL}} \).

---

Would you like more details or have any questions?

Here are 5 follow-up questions you might consider:
1. How do mass percentages influence the density calculation of solutions?
2. Why is the molarity of a solution temperature-dependent?
3. How would the density change if the mass percentage of perchloric acid was higher?
4. What role does molarity play in determining the number of moles in a given volume?
5. How can you derive a formula for dilution from the concept of conservation of moles?

**Tip:** When performing dilutions, remember that the number of moles of solute remains constant before and after dilution!

A perchloric acid solution is 35.00 percent perchloric acid by mass. It is simultaneously 4.4311 M at 25°C. Compute the density of this solution in g cm-3. What volume of this solution should be used in mixing 1.00 L of a 0.63 M perchloric acid solution?

Learn how to calculate the density of a 35% perchloric acid solution with a molarity of 4.4311 M, and determine the volume required to dilute it to 0.63 M. This problem covers key chemistry concepts such as molarity, density, and solution dilution.

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