The density of NaOH solution is `1.2 g cm^(-3)`. The molality of this solution is _______ m.
(Round off to the Nearest Integer)
[Use : Atomic masses : Na : 23.0 u O : 16.0 u H : 1.0 u]
Density of `H_(2)O : 1.0 g cm^(-3)`]

Answer

To determine the molality of the NaOH solution, follow these steps:

1. **Understand the given data:**
- Density of NaOH solution, \( \rho = 1.2 \, \text{g/cm}^3 \)
- Atomic masses: \( \text{Na} = 23.0 \, \text{u} \), \( \text{O} = 16.0 \, \text{u} \), \( \text{H} = 1.0 \, \text{u} \)
- Density of water, \( \rho_{\text{H}_2\text{O}} = 1.0 \, \text{g/cm}^3 \)

2. **Assume a volume of the solution:**
- Let the volume of the solution be \( x \, \text{cm}^3 \).

3. **Calculate the mass of the solution:**
- Using the density formula \( \rho = \frac{\text{mass}}{\text{volume}} \):
\[ \text{Mass of solution} = \rho \times \text{volume} = 1.2 \times x = 1.2x \, \text{g} \]

4. **Calculate the mass of the solvent (water):**
- Given the density of water is \( 1.0 \, \text{g/cm}^3 \):
\[ \text{Mass of water} = 1.0 \times x = x \, \text{g} \]

5. **Calculate the mass of the solute (NaOH):**
- The mass of the solute is the difference between the mass of the solution and the mass of the solvent:
\[ \text{Mass of NaOH} = 1.2x - x = 0.2x \, \text{g} \]

6. **Calculate the molar mass of NaOH:**
- Molar mass of NaOH = \( 23.0 + 16.0 + 1.0 = 40.0 \, \text{g/mol} \)

7. **Calculate the number of moles of NaOH:**
- Number of moles of NaOH = \(\frac{\text{mass of NaOH}}{\text{molar mass of NaOH}}\):
\[ \text{Number of moles of NaOH} = \frac{0.2x}{40.0} = 0.005x \, \text{mol} \]

8. **Calculate the molality of the solution:**
- Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality} = \frac{\text{number of moles of solute}}{\text{mass of solvent in kg}} = \frac{0.005x}{\frac{x}{1000}} = 0.005x \times \frac{1000}{x} = 5 \, \text{mol/kg} \]

9. **Round off to the nearest integer:**
- The molality of the NaOH solution is \( 5 \, \text{mol/kg} \).