Calculate the amount of ethanoic acid in moles and the number of water molecules produced, given the molarity of ethanoic acid solution A is 0.100 mol dm-3.

The molarity of Solution A is given as $0.100 \text{ mol dm}^{-3}$.

$$\text{Moles of CH}_3\text{COOH} = \text{Molarity} \times \text{Volume (in dm}^3)$$ $$\text{Moles of CH}_3\text{COOH} = 0.100 \text{ mol dm}^{-3} \times 0.0250 \text{ dm}^3$$ $$\boxed{\text{Moles of CH}_3\text{COOH} = \textbf{0.00250 mol}}$$

The balanced equation for the reaction is:

$$\text{CH}_3\text{COOH}_{(aq)} + \text{KOH}_{(aq)} \rightarrow \text{CH}_3\text{COOK}_{(aq)} + \text{H}_2\text{O}_{(l)}$$

From the stoichiometry, 1 mole of $\text{CH}_3\text{COOH}$ reacts to produce 1 mole of $\text{H}_2\text{O}$.
Therefore, the moles of water produced are equal to the moles of ethanoic acid reacted:

$$\text{Moles of H}_2\text{O} = 0.00250 \text{ mol}$$

Avogadro's constant ($\text{N}_{\text{A}}$) is given as $6.02 \times 10^{23} \text{ molecules/mol}$.

$$\text{Number of H}_2\text{O molecules} = \text{Moles of H}_2\text{O} \times \text{N}_{\text{A}}$$ $$\text{Number of H}_2\text{O molecules} = 0.00250 \text{ mol} \times 6.02 \times 10^{23} \text{ molecules/mol}$$ $$\text{Number of H}_2\text{O molecules} = 1.505 \times 10^{21} \text{ molecules}$$

Rounding to three significant figures:

$$\boxed{\text{Number of H}_2\text{O molecules} = \textbf{1.51} \times \textbf{10}^{21} \textbf{ molecules}}$$