A 0.5 M aqueous $H_2SO_4$ solution is diluted from 2 litres to 20 litres. The concentration of the resulting solution will be

0.1 N

The correct answer is Option (1) → 0.1 N

Use the dilution formula:

$M_1V_1 = M_2V_2$

Then convert molarity (M) to normality (N) using:

$N = M \times n$

where $n$ = basicity (number of replaceable ${H}^+$). For $\text{H}_2\text{SO}_4$, $\mathbf{n = 2}$.

Step Solution

Initial molarity $M_1 = 0.5 \, M$

Initial volume $V_1 = 2 \, L$

Final volume $V_2 = 20 \, L$

$M_2 = \frac{M_1V_1}{V_2} = \frac{0.5 \times 2}{20} = \frac{1}{20} = 0.05 \, M$

Convert to Normality:

$N = M \times 2 = 0.05 \times 2 = 0.1 \, N$