Concentrated aqueous sulphuric acid is `98% H_(2)SO_(4)` by mass and has a density of `1.84"g cm"^(3)`. What volume of the concentrated acid is required to make 5.0 litre of 0.500 M `H_(2)SO_(4)` solution?

Explanation

To find the volume of concentrated sulfuric acid required to prepare a 5.0 L solution of 0.500 M H₂SO₄, we can follow these steps:

### Step 1: Calculate the number of moles of H₂SO₄ required.
The molarity (M) of a solution is defined as the number of moles of solute per liter of solution.

\[ \text{Moles of } H_2SO_4 = \text{Molarity} \times \text{Volume (L)} \]

Given:
- Molarity = 0.500 M
- Volume = 5.0 L

\[ \text{Moles of } H_2SO_4 = 0.500 \, \text{mol/L} \times 5.0 \, \text{L} = 2.5 \, \text{mol} \]

### Step 2: Calculate the mass of H₂SO₄ needed.
To find the mass, we use the molar mass of H₂SO₄. The molar mass of H₂SO₄ is approximately 98.08 g/mol.

\[ \text{Mass of } H_2SO_4 = \text{Moles} \times \text{Molar Mass} \]

\[ \text{Mass of } H_2SO_4 = 2.5 \, \text{mol} \times 98.08 \, \text{g/mol} = 245.2 \, \text{g} \]

### Step 3: Calculate the mass of concentrated sulfuric acid required.
Since the concentrated sulfuric acid is 98% H₂SO₄ by mass, we can find the mass of the concentrated acid needed.

Let \( x \) be the mass of concentrated sulfuric acid required.

\[ 0.98x = 245.2 \, \text{g} \]

\[ x = \frac{245.2 \, \text{g}}{0.98} \approx 250.2 \, \text{g} \]

### Step 4: Calculate the volume of concentrated sulfuric acid required.
Now, we can use the density of the concentrated sulfuric acid to find the volume. The density is given as 1.84 g/cm³.

\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]

\[ \text{Volume} = \frac{250.2 \, \text{g}}{1.84 \, \text{g/cm}^3} \approx 136.1 \, \text{cm}^3 \]

### Step 5: Convert the volume to liters.
Since 1 cm³ = 0.001 L,

\[ \text{Volume in liters} = 136.1 \, \text{cm}^3 \times 0.001 \, \text{L/cm}^3 \approx 0.1361 \, \text{L} \]

### Final Answer:
The volume of concentrated sulfuric acid required to make 5.0 liters of 0.500 M H₂SO₄ solution is approximately **0.136 L** or **136.1 cm³**.

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