The density (in g `ml^(−1)` ) of a 3.60M sulphuric acid solution having 40% `H_2SO_4`[molar mass =98g `mol^(−1)`] by mass, will be:

Answer

To find the density of a 3.60 M sulfuric acid solution that has 40% H₂SO₄ by mass, we can follow these steps:

### Step 1: Understand the given data
- Molarity (M) of the solution = 3.60 M
- Mass percentage of H₂SO₄ = 40%
- 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 40%, in 100 g of the solution, the mass of H₂SO₄ is:
\[ \text{Mass of H₂SO₄} = 40\% \text{ of } 100 \text{ g} = 40 \text{ g} \]

### Step 3: Calculate the number of moles of H₂SO₄
Using the molar mass of H₂SO₄, we can find the number of moles:
\[ \text{Moles of H₂SO₄} = \frac{\text{Mass of H₂SO₄}}{\text{Molar mass of H₂SO₄}} = \frac{40 \text{ g}}{98 \text{ g/mol}} \approx 0.408 \text{ mol} \]

### Step 4: Relate molarity to volume
Molarity (M) is defined as moles of solute per liter of solution. Therefore, we can express this relationship as:
\[ \text{Molarity} = \frac{\text{Moles of H₂SO₄}}{\text{Volume of solution in L}} \]
Given that the molarity is 3.60 M, we can rearrange this to find the volume of the solution:
\[ 3.60 = \frac{0.408}{\text{Volume in L}} \]
\[ \text{Volume in L} = \frac{0.408}{3.60} \approx 0.1133 \text{ L} = 113.3 \text{ mL} \]

### Step 5: Calculate the mass of the solution
The total mass of the solution is given as 100 g (from our earlier assumption of 100 g of solution).

### Step 6: Calculate the density of the solution
Density (D) is defined as mass per unit volume. Therefore:
\[ D = \frac{\text{Mass of solution}}{\text{Volume of solution in mL}} = \frac{100 \text{ g}}{113.3 \text{ mL}} \approx 0.88 \text{ g/mL} \]

### Final Answer
The density of the 3.60 M sulfuric acid solution is approximately **0.88 g/mL**.