The density of 2M solution of acetic acid (Mol. Wt. 60) is 1.02 g `ml^(-1)`. The molality of the solution 'X' is

Options

A. `1.0`
B. `2.0`
C. `2.22`
D. `2.25`

Explanation

To find the molality of a 2M solution of acetic acid with a density of 1.02 g/ml, we can follow these steps:

### Step 1: Understand the definition of molarity
Molarity (M) is defined as the number of moles of solute per liter of solution. Here, we have a 2M solution, which means there are 2 moles of acetic acid in 1 liter of solution.

### Step 2: Calculate the mass of the solution
Using the density of the solution, we can calculate the mass of 1 liter 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
To find the mass of the solute (acetic acid), we can use the number of moles and the molar mass. The molar mass of acetic acid is given as 60 g/mol.

\[ \text{Mass of solute} = \text{Number of moles} \times \text{Molar mass} = 2 \, \text{moles} \times 60 \, \text{g/mol} = 120 \, \text{g} \]

### Step 4: Calculate the mass of the solvent
Now, we can find the mass of the solvent by subtracting the mass of the solute from the mass of the solution.

\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute} = 1020 \, \text{g} - 120 \, \text{g} = 900 \, \text{g} \]

### 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.

\[ \text{Mass of solvent in kg} = \frac{900 \, \text{g}}{1000} = 0.9 \, \text{kg} \]

### Step 6: Calculate the molality
Finally, we can calculate the molality (m) using the formula:

\[ \text{Molality} (m) = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{2 \, \text{moles}}{0.9 \, \text{kg}} \approx 2.22 \, \text{mol/kg} \]

### Final Answer
Thus, the molality of the solution is approximately **2.22 molal**.

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