What volume of 0.5 mol dm-3 H2SO4 will exactly neutralize 20cm3 of 0.1 mol dm-3 NaOH solution?

To solve this problem, we can use the concept of mole ratios and the equation: H2SO4 + 2NaOH -> Na2SO4 + 2H2O From the equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, to neutralize 1 mole of NaOH, we need 1/2 mole of H2SO4. This means that for the given 0.1 mol dm-3 NaOH solution, we need 0.05 moles of H2SO4 to neutralize it. Now, we can use the equation: moles = concentration x volume (in dm3) to calculate the volume of 0.5 mol dm-3 H2SO4 needed to neutralize 0.05 moles of NaOH. 0.5 x V = 0.05 V = 0.05 / 0.5 V = 0.1 dm3 But the volume of NaOH solution given in the problem is 20 cm3 or 0.02 dm3. Therefore, the volume of H2SO4 required to neutralize it is: V = 0.1 dm3 x (0.02 dm3 / 1 dm3) V = 0.002 dm3 or 2 cm3 Therefore, the answer is option A, 2.0 cm3.