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$^{-3}$]
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To convert molarity to molality:
Assume 1 L of solution, use density to find mass of solution,
subtract solute mass to get solvent mass, then apply the molality formula.
The given question is incomplete in the original paper.
To proceed, we assume that the NaOH solution is 1 Molar,
which is the standard assumption used in such questions. Step 1: Meaning of 1 M solution
A 1 M NaOH solution contains:
\[
1 \text{ mole of NaOH in } 1 \text{ litre (1000 mL) of solution}
\]
Step 2: Mass of solute (NaOH)
\[
\text{Molar mass of NaOH} = 23 + 16 + 1 = 40 \text{ g mol}^{-1}
\]
\[
\text{Mass of NaOH} = 1 \times 40 = 40 \text{ g}
\]
Step 3: Mass of solution
\[
\text{Density of solution} = 1.2 \text{ g mL}^{-1}
\]
\[
\text{Mass of solution} = 1.2 \times 1000 = 1200 \text{ g}
\]
Step 4: Mass of solvent (water)
\[
\text{Mass of solvent} = 1200 - 40 = 1160 \text{ g} = 1.16 \text{ kg}
\]
Step 5: Calculate molality
Molality is defined as:
\[
m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}
\]
\[
m = \frac{1}{1.16} = 0.862 \text{ m}
\]
Step 6: Final Answer
Rounding off to the nearest integer:
\[
\boxed{m = 1}
\]
