`0.1 M NaOH (aq)` solution having density `1.2 gm//ml` will have molality

Options

A. `0.2 M`
B. `0.15 M`
C. `0.08 M`
D. `0.12 M`

Explanation

To find the molality of a 0.1 M NaOH (aq) solution with a density of 1.2 g/mL, we can follow these steps:

### Step 1: Understand the Definitions
- **Molarity (M)** is defined as the number of moles of solute per liter of solution.
- **Molality (m)** is defined as the number of moles of solute per kilogram of solvent.
- **Density** is the mass per unit volume of the solution.

### Step 2: Calculate the Mass of the Solution
Given the density of the solution is 1.2 g/mL, we can calculate the mass of 1 liter (1000 mL) of the solution:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.2 \, \text{g/mL} \times 1000 \, \text{mL} = 1200 \, \text{g} \]

### Step 3: Calculate the Moles of NaOH
Using the molarity of the solution:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
Given that the molarity is 0.1 M, the number of moles of NaOH in 1 liter of solution is:
\[ \text{Moles of NaOH} = 0.1 \, \text{mol/L} \times 1 \, \text{L} = 0.1 \, \text{mol} \]

### Step 4: Calculate the Mass of NaOH
The molar mass of NaOH is approximately 40 g/mol. Therefore, the mass of NaOH in the solution is:
\[ \text{Mass of NaOH} = \text{moles} \times \text{molar mass} = 0.1 \, \text{mol} \times 40 \, \text{g/mol} = 4 \, \text{g} \]

### Step 5: Calculate the Mass of the Solvent (Water)
To find the mass of the solvent (water), we subtract the mass of NaOH from the total mass of the solution:
\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of NaOH} = 1200 \, \text{g} - 4 \, \text{g} = 1196 \, \text{g} \]

### Step 6: Convert the Mass of Solvent to Kilograms
Since molality is expressed in kg of solvent, we convert grams to kilograms:
\[ \text{Mass of solvent in kg} = \frac{1196 \, \text{g}}{1000} = 1.196 \, \text{kg} \]

### Step 7: Calculate Molality
Now we can calculate the molality using the formula:
\[ \text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1 \, \text{mol}}{1.196 \, \text{kg}} \approx 0.0835 \, \text{m} \]

### Step 8: Round the Result
Rounding this to two decimal places gives:
\[ \text{Molality} \approx 0.08 \, \text{m} \]

### Final Answer
The molality of the 0.1 M NaOH solution is approximately **0.08 m**.