If the sea water is 3.5% (w/w) aqueous solution of NaCI then its
molality will be:-

Options

A. 0.598 m
B. 36.27 m
C. 0.62 m
D. 0.578 m

Explanation

To find the molality of a 3.5% (w/w) aqueous solution of NaCl, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The solution is 3.5% (w/w) NaCl, which means there are 3.5 grams of NaCl in 100 grams of the solution.

2. **Finding the Mass of the Solvent**:
- The mass of the solution is 100 grams.
- The mass of the solute (NaCl) is 3.5 grams.
- Therefore, the mass of the solvent (water) can be calculated as:
\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute} = 100 \, \text{g} - 3.5 \, \text{g} = 96.5 \, \text{g} \]

3. **Converting Mass of Solvent to Kilograms**:
- To use the formula for molality, we need the mass of the solvent in kilograms:
\[ \text{Mass of solvent in kg} = \frac{96.5 \, \text{g}}{1000} = 0.0965 \, \text{kg} \]

4. **Calculating the Number of Moles of NaCl**:
- The molar mass of NaCl is approximately 58.5 g/mol.
- The number of moles of NaCl can be calculated using the formula:
\[ \text{Moles of NaCl} = \frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}} = \frac{3.5 \, \text{g}}{58.5 \, \text{g/mol}} \approx 0.0599 \, \text{mol} \]

5. **Calculating Molality**:
- Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.0599 \, \text{mol}}{0.0965 \, \text{kg}} \approx 0.620 \, \text{mol/kg} \]

6. **Final Answer**:
- The molality of the solution is approximately 0.62 mol/kg.

### Final Result:
The molality of the 3.5% (w/w) aqueous solution of NaCl is **0.62 mol/kg**.

---