Question

How many grams of $NaOH$ is required to prepare $5.0$ litre of $0.1\text{ N}$ solution? (Given: At. wt: $H=1$, $O=16$, $Na=23$)

• A. 20
• B. 30
• C. 10
• D. 50

Hint:

For $NaOH$, Normality is equal to Molarity because its n-factor is 1. If you were dealing with $H_2SO_4$, the equivalent weight would be half the molecular weight because the n-factor is 2.

Answer 1

The Correct Option is A

Step 1: Calculate the Molecular Weight and Equivalent Weight: Molecular Weight of $NaOH = 23 (Na) + 16 (O) + 1 (H) = 40\text{ g/mol}$. Since $NaOH$ is a monoacidic base, its acidity is 1. $$\text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{Acidity}} = \frac{40}{1} = 40$$

Step 2: Use the Normality Formula: Normality ($N$) is given by: $$N = \frac{\text{Mass of Solute (w)}}{\text{Equivalent Weight (E)} \times \text{Volume in Litres (V)}}$$

Step 3: Solve for Mass (w): Rearranging the formula: $$w = N \times E \times V$$ Substituting the given values ($N=0.1$, $E=40$, $V=5.0$): $$w = 0.1 \times 40 \times 5.0$$ $$w = 4 \times 5.0 = 20\text{ grams}$$ Thus, 20 grams of $NaOH$ are required to prepare the specified solution.