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 : 23.0u, O : 16.0u, H : 1.0u
Density of H$$_2$$O : 1.0 g cm$$^{-3}$$]

From Molarity ($$M$$) and Solvent Density ($$\rho_0$$):

$$\rho = \frac{M \cdot M_1}{1000} + \rho_0$$

Where:

• $$\rho$$

  = Density of the solution (g/mL)

• $$\rho_0$$

  = Density of the pure solvent (g/mL)

• $$M$$

  = Molarity of the solution (mol/L)

• $$M_1$$

  = Molar mass of the solute (g/mol)

Direct Molality ($$m$$) to Molarity ($$M$$) Relation:

Substitute the solution density formula into the standard relation to isolate the solvent density:

$$m = \frac{M}{\rho_0}$$

Note: This specific linear relationship ($$m = M/\rho_0$$) strictly holds true only for ideal solutions where the volumes of the solute and solvent are perfectly additive ($$V_{\text{solution}} = V_{\text{solvent}} + V_{\text{solute}}$$).

Solving for the NaOH Solution

To find the molality, we first need to use the given density of the solution to calculate the molarity ($$M$$), and then apply the direct relation.

1. Identify the Known Variables

• Solute (NaOH) Molar Mass ($$M_1$$): 40 g/mol (Na = 23, O = 16, H = 1)
• Solvent (Water) Density ($$\rho_0$$): 1.0 g/mL (standard density of pure water)
• Solution Density ($$\rho$$): 1.2 g/cm³ (which is equivalent to 1.2 g/mL)

2. Calculate Molarity ($$M$$)

Use the provided solution density formula:

$$\rho = \frac{M \cdot M_1}{1000} + \rho_0$$

Substitute the known values into the equation:

$$1.2 = \frac{M \cdot 40}{1000} + 1.0$$

Subtract 1.0 from both sides:

$$0.2 = \frac{40M}{1000}$$
$$0.2 = 0.04M$$

Solve for $$M$$:

$$M = \frac{0.2}{0.04}$$
$$M = 5 \text{ mol/L}$$

3. Calculate Molality ($$m$$)

Now, use the direct molality to molarity relation provided in your prompt:

$$m = \frac{M}{\rho_0}$$

Substitute the calculated molarity and the density of water:

$$m = \frac{5}{1.0}$$
$$m = 5 \text{ mol/kg}$$

Final Answer:

The molality of the NaOH solution is 5 m (mol/kg).