The density (in g `ml^(−1)` ) of a 3.60M sulphuric acid solution having 25% `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 25% \( H_2SO_4 \) by mass, we can follow these steps:

### Step 1: Understand the mass percentage
The mass percentage of \( H_2SO_4 \) in the solution is given as 25%. This means that in 100 g of the solution, there are 25 g of \( H_2SO_4 \).

### Step 2: Calculate the moles of \( H_2SO_4 \)
To find the number of moles of \( H_2SO_4 \), we use the formula:
\[ \text{Moles of } H_2SO_4 = \frac{\text{mass of } H_2SO_4}{\text{molar mass of } H_2SO_4} \]
Given that the molar mass of \( H_2SO_4 \) is 98 g/mol, we can calculate:
\[ \text{Moles of } H_2SO_4 = \frac{25 \, \text{g}}{98 \, \text{g/mol}} \approx 0.2551 \, \text{mol} \]

### Step 3: Use the molarity to find the volume of the solution
Molarity (M) is defined as:
\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
We know the molarity is 3.60 M, so we can rearrange the formula to find the volume:
\[ \text{Volume of solution (L)} = \frac{\text{moles of } H_2SO_4}{\text{Molarity}} = \frac{0.2551 \, \text{mol}}{3.60 \, \text{mol/L}} \approx 0.0704 \, \text{L} \]
Convert this volume to milliliters:
\[ \text{Volume of solution (mL)} = 0.0704 \, \text{L} \times 1000 \, \text{mL/L} \approx 70.4 \, \text{mL} \]

### Step 4: 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 initial assumption of 100 g of solution). Thus:
\[ D = \frac{100 \, \text{g}}{70.4 \, \text{mL}} \approx 1.42 \, \text{g/mL} \]

### Final Answer
The density of the 3.60 M sulfuric acid solution is approximately **1.42 g/mL**.

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