Question

A $5$ molar solution of $H_2SO_4$ is diluted from 1 litre to a volume of 10 litres, the normality of the solution will be

• A. 1 N
• B. 0.1 N
• C. 5 N
• D. 0.5 N

Answer 1

The Correct Option is A

To solve this problem, we need to determine the normality of a sulfuric acid (\( H_2SO_4 \)) solution after dilution. Let's go through the steps:

1. Understanding Molarity: The molarity (M) given is \(5\) molar, which means there are \(5\) moles of \( H_2SO_4 \) per liter of solution.
2. Normality of \( H_2SO_4 \): Normality (N) is the number of gram equivalents of solute per liter of solution. For acids, normality is determination based on the number of replaceable hydrogen ions (\( H^+ \)).
3. \( H_2SO_4 \) can dissociate to give two hydrogen ions: H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}
4. Therefore, the normality of a \(5\) molar solution of \( H_2SO_4 \) is: N = 2 \times M = 2 \times 5 = 10 \text{ N}
5. After Dilution: Using the formula for dilution: N_1V_1 = N_2V_2 where:
   • \(N_1 = 10 \text{ N}\) (initial normality)
   • \(V_1 = 1 \text{ L}\) (initial volume)
   • \(V_2 = 10 \text{ L}\) (final volume)
   • \(N_2 = ?\) (final normality)
   Substituting the values, we have: 10 \times 1 = N_2 \times 10
6. Solving for \(N_2\):
7. N_2 = \frac{10}{10} = 1 \text{ N}
8. After diluting a 5 molar \( H_2SO_4 \) solution from 1 liter to 10 liters, the normality becomes \(1 \text{ N}\).

The correct answer is therefore 1 N.