To solve the problem of determining the volume of concentrated sulfuric acid required to prepare 1 liter of a 0.1 M \( H_2SO_4 \) solution, we can follow these steps:
### Step 1: Understand the Given Information
- Concentrated sulfuric acid is 98% \( H_2SO_4 \) by mass.
- Density of the concentrated acid is 1.80 g/mL.
- We need to prepare 1 liter (1000 mL) of a 0.1 M \( H_2SO_4 \) solution.
### Step 2: Calculate the Molarity of Concentrated \( H_2SO_4 \)
1. **Calculate the mass of 100 mL of concentrated \( H_2SO_4 \)**:
\[
\text{Mass} = \text{Volume} \times \text{Density} = 100 \, \text{mL} \times 1.80 \, \text{g/mL} = 180 \, \text{g}
\]
2. **Calculate the mass of \( H_2SO_4 \) in this solution**:
\[
\text{Mass of } H_2SO_4 = 98\% \text{ of } 180 \, \text{g} = 0.98 \times 180 \, \text{g} = 176.4 \, \text{g}
\]
3. **Calculate the number of moles of \( H_2SO_4 \)**:
\[
\text{Molar mass of } H_2SO_4 = 98 \, \text{g/mol}
\]
\[
\text{Moles of } H_2SO_4 = \frac{176.4 \, \text{g}}{98 \, \text{g/mol}} \approx 1.80 \, \text{mol}
\]
4. **Calculate the molarity of the concentrated \( H_2SO_4 \)**:
Since this is the molarity in 100 mL, we convert it to liters:
\[
\text{Molarity (M)} = \frac{\text{Moles}}{\text{Volume in L}} = \frac{1.80 \, \text{mol}}{0.1 \, \text{L}} = 18.0 \, \text{M}
\]
### Step 3: Use the Dilution Equation
We can use the dilution equation \( M_1 V_1 = M_2 V_2 \) to find the volume of concentrated acid needed:
- \( M_1 = 18.0 \, \text{M} \) (molarity of concentrated \( H_2SO_4 \))
- \( M_2 = 0.1 \, \text{M} \) (desired molarity)
- \( V_2 = 1 \, \text{L} = 1000 \, \text{mL} \)
### Step 4: Solve for \( V_1 \)
\[
M_1 V_1 = M_2 V_2
\]
\[
18.0 \, \text{M} \times V_1 = 0.1 \, \text{M} \times 1000 \, \text{mL}
\]
\[
V_1 = \frac{0.1 \times 1000}{18.0} = \frac{100}{18.0} \approx 5.56 \, \text{mL}
\]
### Final Answer
The volume of concentrated sulfuric acid required to make 1 liter of a 0.1 M \( H_2SO_4 \) solution is approximately **5.56 mL**.
