The density (in g `ml^(−1)` ) of a 3.60M sulphuric acid solution having 8.5% `H_2SO_4`[molar mass =98g `mol^(−1)`] by mass, will be:

Explanation

To find the density of a 3.60 M sulfuric acid solution that has 8.5% H₂SO₄ by mass, we can follow these steps:

### Step 1: Understand the mass percentage
The mass percentage of H₂SO₄ is given as 8.5%. This means that in 100 grams of the solution, there are 8.5 grams of H₂SO₄.

### Step 2: Calculate the moles of H₂SO₄
To find the number of moles of H₂SO₄, we can use the formula:
\[ \text{Moles of H₂SO₄} = \frac{\text{mass of H₂SO₄}}{\text{molar mass of H₂SO₄}} \]
Given that the molar mass of H₂SO₄ is 98 g/mol:
\[ \text{Moles of H₂SO₄} = \frac{8.5 \text{ g}}{98 \text{ g/mol}} \approx 0.08673 \text{ mol} \]

### Step 3: Use the molarity to find the volume of the solution
Molarity (M) is defined as moles of solute per liter of solution. We know the molarity is 3.60 M, so we can rearrange the formula to find the volume of the solution:
\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
Rearranging gives:
\[ \text{Volume of solution in liters} = \frac{\text{moles of solute}}{\text{Molarity}} = \frac{0.08673 \text{ mol}}{3.60 \text{ M}} \approx 0.0241 \text{ L} \]
Converting this to milliliters:
\[ \text{Volume of solution in mL} = 0.0241 \text{ L} \times 1000 \approx 24.1 \text{ mL} \]

### Step 4: Calculate the density of the solution
Density (D) is defined as mass per unit volume. The mass of the solution is 100 grams (since we used 100 grams of solution for our calculations). Therefore:
\[ \text{Density} = \frac{\text{mass of solution}}{\text{volume of solution}} = \frac{100 \text{ g}}{24.1 \text{ mL}} \approx 4.15 \text{ g/mL} \]

### Final Answer
The density of the 3.60 M sulfuric acid solution is approximately **4.15 g/mL**.

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