To solve the given problem, we need to follow a series of steps involving the concepts of molarity and dilution. Let's break it down step-by-step:
### Part 1: Calculate the Molar Concentration of the Initial Solution
1. Determine the number of moles of \( H_2SO_4 \):
- We are given the mass of \( H_2SO_4 \): \( 73.5 \) grams.
- The molar mass of \( H_2SO_4 \) (sulfuric acid) is approximately \( 98.079 \) g/mol.
- To find the number of moles, we use the formula:
[tex]\[
\text{moles} = \frac{\text{mass}}{\text{molar mass}}
\][/tex]
Plugging in the values:
[tex]\[
\text{moles} = \frac{73.5 \, \text{g}}{98.079 \, \text{g/mol}} \approx 0.7494 \, \text{mol}
\][/tex]
2. Calculate the initial molar concentration:
- The volume of the solution is given as \( 3.0 \) liters.
- The molarity (concentration) of a solution is calculated using the formula:
[tex]\[
\text{Molarity (M)} = \frac{\text{moles}}{\text{volume (L)}}
\][/tex]
Plugging in the values:
[tex]\[
\text{Molarity (M)} = \frac{0.7494 \, \text{mol}}{3.0 \, \text{L}} \approx 0.2498 \, \text{M}
\][/tex]
### Part 2: Determine the Volume of Water Required to Dilute the Solution to 0.18 M
1. Set up the dilution equation:
- The dilution equation is given by \( C_1V_1 = C_2V_2 \), where:
- \( C_1 \) is the initial concentration
- \( V_1 \) is the initial volume
- \( C_2 \) is the final concentration
- \( V_2 \) is the final volume
- We have:
- \( C_1 = 0.2498 \) M
- \( V_1 = 3.0 \) L
- \( C_2 = 0.18 \) M
2. Solve for \( V_2 \):
[tex]\[
0.2498 \times 3.0 = 0.18 \times V_2
\][/tex]
[tex]\[
V_2 = \frac{0.2498 \times 3.0}{0.18} \approx 4.1633 \, \text{L}
\][/tex]
3. Calculate the volume of water needed:
- The final volume \( V_2 \) is \( 4.1633 \) L.
- The initial volume is \( 3.0 \) L.
- The volume of water to be added is:
[tex]\[
\text{Volume of water added} = V_2 - V_1 = 4.1633 \, \text{L} - 3.0 \, \text{L} \approx 1.1633 \, \text{L}
\][/tex]
### Summary
- Molar concentration of the initial solution: \( 0.2498 \) M
- Volume of water needed to dilute the solution to [tex]\( 0.18 \)[/tex] M: approximately [tex]\( 1.1633 \)[/tex] L
