A piece of aluminium weighing 2.7 g is heated with 75 mL of `H_(2)SO_(4)` which has a density of `1.18"g mL"^(-1)` and contains `24.7%` by mass. When whole of the metal had dissolved, the solution was diluted to 400 mL. Calculate the molarity of free `H_(2)SO_(4)` in the resulting solution.
Hence, molarity `=(7.17)/(98)xx(1)/(400)xx1000=0.183M`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the mass of H₂SO₄ in the solution
Given that the solution contains 24.7% H₂SO₄ by mass, we can find the mass of H₂SO₄ in 75 mL of the solution.

1. **Calculate the mass of the solution:**
\[ \text{Mass of solution} = \text{Volume} \times \text{Density} = 75 \, \text{mL} \times 1.18 \, \text{g/mL} = 88.5 \, \text{g} \]

2. **Calculate the mass of H₂SO₄:**
\[ \text{Mass of H₂SO₄} = \text{Mass of solution} \times \frac{24.7}{100} = 88.5 \, \text{g} \times 0.247 = 21.85 \, \text{g} \]

### Step 2: Calculate the number of moles of H₂SO₄
To find the number of moles of H₂SO₄, we use its molar mass. The molar mass of H₂SO₄ is approximately 98 g/mol.

\[ \text{Moles of H₂SO₄} = \frac{\text{Mass of H₂SO₄}}{\text{Molar mass of H₂SO₄}} = \frac{21.85 \, \text{g}}{98 \, \text{g/mol}} \approx 0.222 \, \text{mol} \]

### Step 3: Determine the total volume of the solution after dilution
The solution is diluted to a total volume of 400 mL.

### Step 4: Calculate the molarity of H₂SO₄ in the diluted solution
Molarity (M) is defined as the number of moles of solute divided by the volume of solution in liters.

1. **Convert the volume from mL to L:**
\[ \text{Volume in L} = \frac{400 \, \text{mL}}{1000} = 0.400 \, \text{L} \]

2. **Calculate the molarity:**
\[ \text{Molarity} = \frac{\text{Moles of H₂SO₄}}{\text{Volume in L}} = \frac{0.222 \, \text{mol}}{0.400 \, \text{L}} = 0.555 \, \text{M} \]

### Final Result
The molarity of free H₂SO₄ in the resulting solution is approximately **0.555 M**.

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