The density (in g `ml^(−1)` ) of a 3.60M sulphuric acid solution having 33% `H_2SO_4`[molar mass =98g `mol^(−1)`] by mass, will be:

Options

A. 1.07g/ml
B. 2.4g/ml
C. 1.22g/ml
D. None of these

Explanation

To find the density of a 3.60 M sulfuric acid solution that has 33% H₂SO₄ by mass, we can follow these steps:

### Step 1: Understand the given information
- Molarity (M) = 3.60 M
- Mass percentage of H₂SO₄ = 33%
- Molar mass of H₂SO₄ = 98 g/mol

### Step 2: Calculate the mass of H₂SO₄ in 100 g of solution
Since the mass percentage of H₂SO₄ is 33%, in 100 g of the solution, the mass of H₂SO₄ is:
\[ \text{Mass of H₂SO₄} = 33\% \text{ of } 100 \text{ g} = 33 \text{ g} \]

### Step 3: Calculate the number of moles of H₂SO₄
Using the molar mass of H₂SO₄, we can find the number of moles:
\[ \text{Moles of H₂SO₄} = \frac{\text{Mass of H₂SO₄}}{\text{Molar mass of H₂SO₄}} = \frac{33 \text{ g}}{98 \text{ g/mol}} \approx 0.3367 \text{ mol} \]

### Step 4: Use the molarity to find the volume of the solution
Molarity (M) is defined as moles of solute per liter of solution. Therefore, 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:
\[ \text{Volume of solution} = \frac{\text{Moles of solute}}{\text{Molarity}} = \frac{0.3367 \text{ mol}}{3.60 \text{ mol/L}} \approx 0.0934 \text{ L} = 93.4 \text{ mL} \]

### Step 5: Calculate the density of the solution
Density is defined as mass per unit volume. The mass of the solution is 100 g (as we considered 100 g of the solution), and the volume is 93.4 mL. Therefore, the density (D) can be calculated as:
\[ D = \frac{\text{Mass of solution}}{\text{Volume of solution}} = \frac{100 \text{ g}}{93.4 \text{ mL}} \approx 1.07 \text{ g/mL} \]

### Final Answer
The density of the 3.60 M sulfuric acid solution is approximately **1.07 g/mL**.

---