The neutralization occurs when 10 mL of 0.1 M acid 'A' is allowed to react with 30 mL of 0.05 M base M(OH)2. The…

Step 1 — At neutralisation, equivalents of acid = equivalents of base:

$n_{\mathrm{acid}} \times \text{basicity} = n_{\mathrm{base}} \times \text{acidity}$

Step 2 — Moles of each: $n_{\mathrm{acid}} = 0.1 \times 10 \times 10^{-3} = 1 \times 10^{-3}\ \mathrm{mol}$ $n_{\mathrm{base}} = 0.05 \times 30 \times 10^{-3} = 1.5 \times 10^{-3}\ \mathrm{mol}$

Step 3 — Acidity of $\mathrm{M(OH)_2}$: $\mathrm{M(OH)_2}$ provides 2 $\mathrm{OH^-}$ per formula unit → acidity = 2

Step 4 — Equivalents: $n_{\mathrm{acid}} \times \text{basicity} = n_{\mathrm{base}} \times 2$ $1 \times 10^{-3} \times \text{basicity} = 1.5 \times 10^{-3} \times 2$ $\text{basicity} = \frac{3 \times 10^{-3}}{1 \times 10^{-3}} = 3$

Answer: Option (3) — 3