Understanding the Concept:
Normality (\(\text{N}\)) represents the total number of gram equivalents of solute dissolved per liter of solution. Molarity (\(\text{M}\)) represents the total moles of solute per liter of solution.
The core mathematical expressions needed are:
\[
\text{Number of Moles} = \frac{\text{Given Mass (g)}}{\text{Molecular Weight (g/mol)}}
\]
\[
\text{Normality (N)} = \frac{\text{Mass of solute (g)}}{\text{Equivalent Weight}} \times \frac{1000}{\text{Volume of solution (mL)}}
\]
\[
\text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{Acidity of Base}}
\]
Step 1: Calculate the Molecular Weight and Equivalent Weight of \(\text{NaOH}\)
Atomic masses are roughly: \(\text{Na} = 23\), \(\text{O} = 16\), \(\text{H} = 1\).
\[
\text{Molecular Weight of NaOH} = 23 + 16 + 1 = 40\text{ g/mol}
\]
Since Sodium Hydroxide (\(\text{NaOH}\)) is a monoacidic base, it releases exactly one hydroxyl ion (\(\text{OH}^-\)) per formula unit. Its acidity factor is 1.
\[
\text{Equivalent Weight of NaOH} = \frac{40}{1} = 40\text{ g/eq}
\]
Step 2: Compute Normality (\(\text{N}\))
Substitute the given constraints (Mass = 20 g, Volume = 500 mL) into the normality equation:
\[
\text{Normality} = \frac{20\text{ g}}{40\text{ g/eq}} \times \frac{1000}{500\text{ mL}}
\]
Simplify the individual structural fractions:
\[
\text{Normality} = 0.5 \times 2
\]
\[
\text{Normality} = 1\text{ N}
\]
Step 3: Verify Molarity status relative to Option C
Since the chemical valence factor is 1, Normality matches Molarity directly (\(\text{N} = \text{M} \times 1\)). Hence, the solution is also equal to 1 M. Option C states 0.5 M, which is mathematically incorrect. Therefore, Option B (1 N) stands as the accurate answer choice.