Pls let there be 1st 2nd 3rd reading

Here is the completed table with the assumed readings:

Titration	Final Burette Reading ($\text{cm}^3$)	Initial Burette Reading ($\text{cm}^3$)	Volume of A used ($\text{cm}^3$)
Rough	22.50	0.00	22.50
1st	22.15	0.00	22.15
2nd	22.17	0.00	22.17
3rd	22.30	0.00	22.30

---

The concordant titres are the 1st and 2nd readings ($22.15 \text{ cm}^3$ and $22.17 \text{ cm}^3$) as they are within $0.10 \text{ cm}^3$ of each other. The 3rd reading ($22.30 \text{ cm}^3$) is not concordant.

$$\text{Average volume of A} = \frac{\text{22.15 cm}^3 + \text{22.17 cm}^3}{2}$$ $$\text{Average volume of A} = \frac{\text{44.32 cm}^3}{2}$$ $$\text{Average volume of A} = \boxed{\text{22.16 cm}^3}$$

---

The reaction between ethanoic acid ($\text{CH}_3\text{COOH}$) and potassium hydroxide ($\text{KOH}$) is:

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

The mole ratio of acid to base ($n_A : n_B$) is $1:1$.

Given:
Concentration of B ($C_B$) = $0.05 \text{ mol dm}^{-3}$
Volume of B ($V_B$) = $25.00 \text{ cm}^3 = 0.025 \text{ dm}^3$
Average volume of A ($V_A$) = $22.16 \text{ cm}^3 = 0.02216 \text{ dm}^3$

Using the titration formula:

$$\frac{C_A V_A}{C_B V_B} = \frac{n_A}{n_B}$$ $$\frac{C_A \times 0.02216 \text{ dm}^3}{0.05 \text{ mol dm}^{-3} \times 0.025 \text{ dm}^3} = \frac{1}{1}$$ $$C_A = \frac{0.05 \text{ mol dm}^{-3} \times 0.025 \text{ dm}^3}{0.02216 \text{ dm}^3}$$ $$C_A = \frac{0.00125}{0.02216} \text{ mol dm}^{-3}$$ $$C_A = \boxed{\text{0.0564 mol dm}^{-3}}$$

---

Amount in moles = Concentration $\times$ Volume

$$\text{Moles of A} = C_A \times V_A$$ $$\text{Moles of A} = 0.0564 \text{ mol dm}^{-3} \times 0.02216 \text{ dm}^3$$ $$\text{Moles of A} = \boxed{0.00125 \text{ mol}}$$

---

From the balanced equation, 1 mole of $\text{CH}_3\text{COOH}$ reacts with 1 mole of $\text{KOH}$ to produce 1 mole of $\text{H}_2\text{O}$.
Therefore, the moles of water formed are equal to the moles of ethanoic acid reacted.
Moles of $\text{H}_2\text{O}$ formed = $0.00125 \text{ mol}$

To find the number of water molecules, multiply by Avogadro's number ($6.02 \times 10^{23} \text{ molecules/mol}$).

$$\text{Number of water molecules} = \text{Moles of H}_2\text{O} \times \text{Avogadro's number}$$ $$\text{Number of water molecules} = 0.00125 \text{ mol} \times 6.02 \times 10^{23} \text{ molecules/mol}$$ $$\text{Number of water molecules} = \boxed{7.525 \times 10^{20} \text{ molecules}}$$