The density (in g `ml^(−1)` ) of a 3.60M sulphuric acid solution having 3.5% `H_2SO_4`[molar mass =98g `mol^(−1)`] by mass, will be:

Options

A. 10g/ml
B. 20g/ml
C. 30g/ml
D. 40g/ml

Explanation

To find the density of a 3.60 M sulfuric acid solution that has a mass percentage of 3.5% H₂SO₄, we can follow these steps:

### Step 1: Understand the mass percentage
The mass percentage of H₂SO₄ is given as 3.5%. This means that in 100 grams of the solution, there are 3.5 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:
- Mass of H₂SO₄ = 3.5 g
- Molar mass of H₂SO₄ = 98 g/mol

Calculating the moles:
\[ \text{Moles of H₂SO₄} = \frac{3.5 \, \text{g}}{98 \, \text{g/mol}} = 0.0357 \, \text{mol} \]

### Step 3: Relate moles to molarity
Molarity (M) is defined as:
\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
Given that the molarity is 3.60 M, we can set up the equation:
\[ 3.60 = \frac{0.0357}{V} \]
Where \( V \) is the volume of the solution in liters.

### Step 4: Solve for volume
Rearranging the equation to find \( V \):
\[ V = \frac{0.0357}{3.60} = 0.00992 \, \text{L} = 9.92 \, \text{ml} \]

### 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 the mass percentage definition). Therefore:
\[ D = \frac{100 \, \text{g}}{9.92 \, \text{ml}} \approx 10.08 \, \text{g/ml} \]

### Step 6: Round to appropriate significant figures
The density can be rounded to 10 g/ml.

### Final Answer
The density of the 3.60 M sulfuric acid solution is approximately **10 g/ml**.

---