Calculate the volume of `80%H_(2)SO_(4)` (density = 1.80 g/cc) required to prepare on litre of `20% H_(2)SO_(4)` (density = 1.25 g/cc).

Explanation

To solve the problem of calculating the volume of 80% H₂SO₄ required to prepare 1 liter of 20% H₂SO₄, we will follow these steps:

### Step 1: Calculate the Molarity of 80% H₂SO₄
1. **Determine the mass of H₂SO₄ in 100 g of solution:**
- Since it is an 80% solution, in 100 g of the solution, there are 80 g of H₂SO₄.

2. **Calculate the number of moles of H₂SO₄:**
- Molar mass of H₂SO₄ = 98 g/mol
- Number of moles = mass (g) / molar mass (g/mol) = 80 g / 98 g/mol = 0.8163 mol

3. **Calculate the volume of the 80% solution:**
- Volume = mass / density = 100 g / 1.80 g/cm³ = 55.56 cm³ = 0.05556 L

4. **Calculate the molarity (M) of the 80% solution:**
- Molarity (M) = moles of solute / volume of solution in liters = 0.8163 mol / 0.05556 L = 14.69 M

### Step 2: Calculate the Molarity of 20% H₂SO₄
1. **Determine the mass of H₂SO₄ in 100 g of solution:**
- In a 20% solution, in 100 g of the solution, there are 20 g of H₂SO₄.

2. **Calculate the number of moles of H₂SO₄:**
- Number of moles = mass (g) / molar mass (g/mol) = 20 g / 98 g/mol = 0.2041 mol

3. **Calculate the volume of the 20% solution:**
- Volume = mass / density = 100 g / 1.25 g/cm³ = 80 cm³ = 0.08 L

4. **Calculate the molarity (M) of the 20% solution:**
- Molarity (M) = moles of solute / volume of solution in liters = 0.2041 mol / 0.08 L = 2.55 M

### Step 3: Use Molarity to Find Volume of 80% H₂SO₄ Required
1. **Use the formula M₁V₁ = M₂V₂:**
- Let M₁ = 14.69 M (molarity of 80% solution)
- Let V₁ = volume of 80% solution required (unknown)
- Let M₂ = 2.55 M (molarity of 20% solution)
- Let V₂ = 1 L (volume of 20% solution)

2. **Set up the equation:**
- 14.69 M * V₁ = 2.55 M * 1 L

3. **Solve for V₁:**
- V₁ = (2.55 M * 1 L) / 14.69 M
- V₁ = 0.1736 L = 173.6 mL

### Final Answer:
The volume of 80% H₂SO₄ required to prepare 1 liter of 20% H₂SO₄ is **173.6 mL**.

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