Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is:

Options

A. `1.14 mol kg^(-1)`
B. `3.28 mol kg^(-1)`
C. `2.28 mol K^(-1)`
D. `0.44 mol kg^(-1)`

Explanation

To find the molality of a 2.05 M solution of acetic acid in water with a density of 1.02 g/mL, we can follow these steps:

### Step 1: Calculate the number of moles of solute (acetic acid)
Given that the molarity (M) of the solution is 2.05 M, this means there are 2.05 moles of acetic acid in 1 liter (1000 mL) of solution.

\[ \text{Number of moles of acetic acid} = 2.05 \text{ moles} \]

### Step 2: Calculate the mass of the solution
Using the density of the solution, we can find the mass of the solution. The density is given as 1.02 g/mL.

\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.02 \text{ g/mL} \times 1000 \text{ mL} = 1020 \text{ g} \]

### Step 3: Calculate the mass of the solute (acetic acid)
To find the mass of the solute, we need the molar mass of acetic acid, which is approximately 60 g/mol.

\[ \text{Mass of acetic acid} = \text{Number of moles} \times \text{Molar mass} = 2.05 \text{ moles} \times 60 \text{ g/mol} = 123 \text{ g} \]

### Step 4: Calculate the mass of the solvent (water)
The mass of the solvent can be calculated by subtracting the mass of the solute from the total mass of the solution.

\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute} = 1020 \text{ g} - 123 \text{ g} = 897 \text{ g} \]

### Step 5: Convert the mass of the solvent to kilograms
Since molality is defined as moles of solute per kilogram of solvent, we need to convert the mass of the solvent from grams to kilograms.

\[ \text{Mass of solvent in kg} = \frac{897 \text{ g}}{1000} = 0.897 \text{ kg} \]

### Step 6: Calculate the molality of the solution
Now we can calculate the molality (m) using the formula:

\[ \text{Molality} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} = \frac{2.05 \text{ moles}}{0.897 \text{ kg}} \approx 2.285 \text{ mol/kg} \]

### Final Answer
The molality of the solution is approximately **2.285 mol/kg**.

---