Find out the volume of 98% w/w `H_(2)SO_(4)` (density = 1.8 gm/ml), must be diluted to prepare 2.6 litres of 2.0 M sulphuric acid solution

Answer

To find the volume of 98% w/w H₂SO₄ needed to prepare 2.6 liters of a 2.0 M sulfuric acid solution, we can follow these steps:

### Step 1: Calculate the number of moles of H₂SO₄ required for the desired solution.
To find the number of moles (n) required, we use the formula:
\[ n = M \times V \]
Where:
- \( M \) = molarity of the solution (2.0 M)
- \( V \) = volume of the solution in liters (2.6 L)

Calculating:
\[ n = 2.0 \, \text{mol/L} \times 2.6 \, \text{L} = 5.2 \, \text{mol} \]

### Step 2: Calculate the mass of H₂SO₄ required.
Using the molar mass of H₂SO₄ (which is approximately 98 g/mol):
\[ \text{mass} = n \times \text{molar mass} \]
\[ \text{mass} = 5.2 \, \text{mol} \times 98 \, \text{g/mol} = 509.6 \, \text{g} \]

### Step 3: Determine the mass of the 98% w/w H₂SO₄ solution needed.
Since the solution is 98% w/w, this means that in 100 g of solution, there are 98 g of H₂SO₄. We can set up a proportion to find the total mass of the solution needed to obtain 509.6 g of H₂SO₄:
\[ \text{mass of solution} = \frac{\text{mass of H₂SO₄}}{0.98} \]
\[ \text{mass of solution} = \frac{509.6 \, \text{g}}{0.98} \approx 520.4 \, \text{g} \]

### Step 4: Calculate the volume of the 98% H₂SO₄ solution using its density.
We know the density of the 98% H₂SO₄ solution is 1.8 g/mL. We can find the volume (V) using the formula:
\[ \text{density} = \frac{\text{mass}}{\text{volume}} \]
Rearranging gives:
\[ \text{volume} = \frac{\text{mass}}{\text{density}} \]
Calculating:
\[ V = \frac{520.4 \, \text{g}}{1.8 \, \text{g/mL}} \approx 289.1 \, \text{mL} \]

### Step 5: Convert the volume from mL to L.
To convert mL to L, we divide by 1000:
\[ V = \frac{289.1 \, \text{mL}}{1000} \approx 0.2891 \, \text{L} \]

### Final Answer:
The volume of 98% w/w H₂SO₄ needed is approximately **289.1 mL** or **0.2891 L**.

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