Calculate the concentration of acid after adding water for each row.

The initial concentration of sulphuric acid is $2 \text{ M}$. The total volume of the solution after adding water is always $50 \text{ cm}^3$. We use the dilution formula $M_1V_1 = M_2V_2$, where $M_1 = 2 \text{ M}$ and $V_2 = 50 \text{ cm}^3$.

• For Volume of 2M H₂SO₄ = 30 cm³: $$M_2 = \frac{M_1V_1}{V_2} = \frac{2 \text{ M} \times 30 \text{ cm}^3}{50 \text{ cm}^3} = \frac{60}{50} = 1.2 \text{ mol dm}^{-3}$$
• For Volume of 2M H₂SO₄ = 20 cm³: $$M_2 = \frac{M_1V_1}{V_2} = \frac{2 \text{ M} \times 20 \text{ cm}^3}{50 \text{ cm}^3} = \frac{40}{50} = 0.8 \text{ mol dm}^{-3}$$
• For Volume of 2M H₂SO₄ = 10 cm³: $$M_2 = \frac{M_1V_1}{V_2} = \frac{2 \text{ M} \times 10 \text{ cm}^3}{50 \text{ cm}^3} = \frac{20}{50} = 0.4 \text{ mol dm}^{-3}$$

• For Time taken = 10.0 s: $$\frac{1}{10.0 \text{ s}} \times 1000 = 100.0 \text{ s}^{-1}$$
• For Time taken = 15.0 s: $$\frac{1}{15.0 \text{ s}} \times 1000 = 66.66... \approx 66.7 \text{ s}^{-1}$$
• For Time taken = 25.0 s: $$\frac{1}{25.0 \text{ s}} \times 1000 = 40.0 \text{ s}^{-1}$$

The completed table is: