The density (in g `ml^(−1)` ) of a 3.60M sulphuric acid solution having 35% `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 contains 35% H₂SO₄ by mass, we can follow these steps:

### Step 1: Understand the given information
We know:
- Molarity (M) = 3.60 M
- Mass percentage of H₂SO₄ = 35%
- 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 35%, in 100 g of the solution, the mass of H₂SO₄ is:
\[ \text{Mass of H₂SO₄} = 35\% \text{ of } 100 \text{ g} = 35 \text{ g} \]

### Step 3: Calculate the number of moles of H₂SO₄
Using the molar mass of H₂SO₄, we can calculate the number of moles:
\[ \text{Moles of H₂SO₄} = \frac{\text{Mass of H₂SO₄}}{\text{Molar mass of H₂SO₄}} = \frac{35 \text{ g}}{98 \text{ g/mol}} = 0.357 \text{ mol} \]

### Step 4: Use the molarity formula to find the volume of the solution
Molarity (M) is defined as:
\[ M = \frac{\text{moles of solute}}{\text{volume of solution in L}} \]
Rearranging this gives:
\[ \text{Volume of solution in L} = \frac{\text{moles of solute}}{M} \]
Substituting the values we have:
\[ \text{Volume of solution in L} = \frac{0.357 \text{ mol}}{3.60 \text{ M}} = 0.0992 \text{ L} \]
Converting this to mL:
\[ \text{Volume of solution in mL} = 0.0992 \text{ L} \times 1000 = 99.2 \text{ mL} \]

### Step 5: Calculate the density of the solution
Density (D) is defined as:
\[ D = \frac{\text{mass of solution}}{\text{volume of solution}} \]
The mass of the solution is 100 g (from our earlier calculation). Therefore:
\[ D = \frac{100 \text{ g}}{99.2 \text{ mL}} \approx 1.01 \text{ g/mL} \]

### Final Answer
The density of the 3.60 M sulfuric acid solution is approximately **1.01 g/mL**.

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