The density (in g `ml^(−1)` ) of a 3.60M sulphuric acid solution having 5.5% `H_2SO_4`[molar mass =98g `mol^(−1)`] by mass, will be:

Options

A. 1.4g/ml
B. 6.4g/ml
C. 4.4g/ml
D. 5.4g/ml

Explanation

To find the density of a 3.60 M sulfuric acid solution that has 5.5% \( H_2SO_4 \) by mass, we can follow these steps:

### Step 1: Understand the given data
- Molarity (M) = 3.60 M
- Mass percentage of \( H_2SO_4 \) = 5.5%
- Molar mass of \( H_2SO_4 \) = 98 g/mol

### Step 2: Calculate the mass of \( H_2SO_4 \) in 100 g of solution
Since the mass percentage of \( H_2SO_4 \) is 5.5%, in 100 g of the solution, the mass of \( H_2SO_4 \) is:
\[ \text{Mass of } H_2SO_4 = 5.5 \text{ g} \]

### Step 3: Calculate the number of moles of \( H_2SO_4 \)
Using the molar mass of \( H_2SO_4 \):
\[ \text{Moles of } H_2SO_4 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{5.5 \text{ g}}{98 \text{ g/mol}} \approx 0.0561 \text{ mol} \]

### Step 4: Relate moles to molarity
Molarity (M) is defined as:
\[ M = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \]
Given that the molarity is 3.60 M, we can set up the equation:
\[ 3.60 = \frac{0.0561}{V} \]
Where \( V \) is the volume of the solution in liters. Rearranging gives:
\[ V = \frac{0.0561}{3.60} \approx 0.0156 \text{ L} = 15.6 \text{ mL} \]

### Step 5: Calculate the mass of the solution
The total mass of the solution is the mass of \( H_2SO_4 \) plus the mass of the solvent (water). Since we have 5.5 g of \( H_2SO_4 \) in 100 g of solution, the mass of the solution is:
\[ \text{Mass of solution} = 100 \text{ g} \]

### Step 6: Calculate the density of the solution
Density (\( D \)) is defined as:
\[ D = \frac{\text{Mass of solution}}{\text{Volume of solution}} \]
Using the mass of the solution (100 g) and the volume (15.6 mL):
\[ D = \frac{100 \text{ g}}{15.6 \text{ mL}} \approx 6.41 \text{ g/mL} \]

### Final Answer
The density of the 3.60 M sulfuric acid solution is approximately **6.41 g/mL**.

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