4 L of 0.02 M aqueous solution of NaCl was diluted by adding 1 L of water. The molality of the resultant solution is……..

Options

A. `0.004`
B. `0.008`
C. `0.012`
D. `0.016`

Explanation

To find the molality of the resultant solution after diluting 4 L of a 0.02 M aqueous solution of NaCl by adding 1 L of water, we can follow these steps:

### Step 1: Calculate the number of moles of NaCl in the original solution.
The number of moles (n) can be calculated using the formula:
\[ n = M \times V \]
Where:
- \( M \) = molarity of the solution (0.02 M)
- \( V \) = volume of the solution in liters (4 L)

Substituting the values:
\[ n = 0.02 \, \text{mol/L} \times 4 \, \text{L} = 0.08 \, \text{mol} \]

### Step 2: Determine the total volume of the diluted solution.
After adding 1 L of water to the original 4 L solution, the total volume (V_total) becomes:
\[ V_{\text{total}} = 4 \, \text{L} + 1 \, \text{L} = 5 \, \text{L} \]

### Step 3: Calculate the new molarity of the solution.
Using the formula for molarity:
\[ M = \frac{n}{V} \]
Where:
- \( n \) = number of moles (0.08 mol)
- \( V \) = total volume of the solution in liters (5 L)

Substituting the values:
\[ M = \frac{0.08 \, \text{mol}}{5 \, \text{L}} = 0.016 \, \text{M} \]

### Step 4: Calculate the molality of the solution.
Molality (m) is defined as the number of moles of solute per kilogram of solvent. Since we have 0.08 moles of NaCl and we added 1 L of water (which has a mass of approximately 1 kg), we can calculate molality as follows:
\[ m = \frac{n}{\text{mass of solvent in kg}} = \frac{0.08 \, \text{mol}}{1 \, \text{kg}} = 0.08 \, \text{mol/kg} \]

### Final Answer:
The molality of the resultant solution is **0.08 mol/kg**.

---