Battery acid is `4.27M H_(2)SO_(4)`, (aq) and has the density of 1.25 g `mL^(-1)`. The molality of `H_(2)SO_(4)`, in the solution is

Options

A. 3.416 m
B. 3.342 m
C. 5.135 m
D. 2.135 m

Explanation

To find the molality of the `H2SO4` solution, we will follow these steps:

### Step 1: Understand the given data
- Molarity (M) of `H2SO4` = 4.27 M
- Density of the solution = 1.25 g/mL

### Step 2: Calculate the mass of the solution
Since molarity is defined as moles of solute per liter of solution, we know that:
- 4.27 M means there are 4.27 moles of `H2SO4` in 1 liter (1000 mL) of solution.

Now, we can calculate the mass of the solution using the density:
\[ \text{Mass of solution} = \text{Volume} \times \text{Density} = 1000 \, \text{mL} \times 1.25 \, \text{g/mL} = 1250 \, \text{g} \]

### Step 3: Calculate the mass of the solute
To find the mass of the solute (`H2SO4`), we need its molar mass. The molar mass of `H2SO4` is calculated as follows:
- Molar mass of `H`: 1 g/mol (2 H = 2 g)
- Molar mass of `S`: 32 g/mol
- Molar mass of `O`: 16 g/mol (4 O = 64 g)

Adding these together:
\[ \text{Molar mass of } H2SO4 = 2 + 32 + 64 = 98 \, \text{g/mol} \]

Now, we can calculate the mass of the solute:
\[ \text{Mass of solute} = \text{Moles of solute} \times \text{Molar mass of solute} = 4.27 \, \text{moles} \times 98 \, \text{g/mol} = 418.46 \, \text{g} \]

### Step 4: Calculate the mass of the solvent
The mass of the solvent can be found by subtracting the mass of the solute from the mass of the solution:
\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute} = 1250 \, \text{g} - 418.46 \, \text{g} = 831.54 \, \text{g} \]

### Step 5: Convert the mass of the solvent to kg
Since molality is expressed in terms of kg of solvent:
\[ \text{Mass of solvent in kg} = \frac{831.54 \, \text{g}}{1000} = 0.83154 \, \text{kg} \]

### Step 6: Calculate the molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{4.27 \, \text{moles}}{0.83154 \, \text{kg}} \approx 5.135 \, \text{mol/kg} \]

### Final Answer
The molality of the `H2SO4` solution is approximately **5.135 mol/kg**.

---