The density of a 2.05 molar solution of acetic acid in water is 1.82g/ml. The molality of the solution is

Options

A. 1.14 mol/kg
B. 3.28 mol/kg
C. 2.28 mol/kg
D. 0.44mol/kg

Explanation

To find the molality of a 2.05 molar solution of acetic acid in water with a density of 1.82 g/mL, we can follow these steps:

### Step 1: Calculate the mass of the solution
Given that the density of the solution is 1.82 g/mL, we can find the mass of 1 liter (1000 mL) of the solution.

\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.82 \, \text{g/mL} \times 1000 \, \text{mL} = 1820 \, \text{g} \]

### Step 2: Calculate the mass of the solute (acetic acid)
We know the molarity (M) of the solution is 2.05 mol/L. This means there are 2.05 moles of acetic acid in 1 liter of solution.

To find the mass of acetic acid, we need its molar mass. The molecular formula of acetic acid (CH₃COOH) gives us:
- Carbon (C): 12 g/mol (2 atoms)
- Hydrogen (H): 1 g/mol (4 atoms)
- Oxygen (O): 16 g/mol (2 atoms)

Calculating the molar mass:
\[ \text{Molar mass of acetic acid} = (2 \times 12) + (4 \times 1) + (2 \times 16) = 24 + 4 + 32 = 60 \, \text{g/mol} \]

Now, we can calculate the mass of acetic acid:
\[ \text{Mass of acetic acid} = \text{Number of moles} \times \text{Molar mass} = 2.05 \, \text{mol} \times 60 \, \text{g/mol} = 123 \, \text{g} \]

### Step 3: Calculate the mass of the solvent (water)
The mass of the solvent can be found by subtracting the mass of the solute from the total mass of the solution:
\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute} = 1820 \, \text{g} - 123 \, \text{g} = 1697 \, \text{g} \]

### Step 4: Convert the mass of the solvent to kilograms
To calculate molality, we need the mass of the solvent in kilograms:
\[ \text{Mass of solvent in kg} = \frac{1697 \, \text{g}}{1000} = 1.697 \, \text{kg} \]

### Step 5: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} = \frac{2.05 \, \text{mol}}{1.697 \, \text{kg}} \approx 1.21 \, \text{mol/kg} \]

### Final Answer
The molality of the solution is approximately **1.21 mol/kg**.

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