If 20 cm3 of sodium hydroxide was neutralized by 20 cm3 of 0.01 mol dm3 tetraoxosulphate(VI) acid, what is the concentration of the solution?

If 20 cm3 of sodium hydroxide was neutralized by 20 cm3 of 0.01 mol dm3 tetraoxosulphate(VI) acid, what is the concentration of the solution?
A. 0.010
B. 0.020
C. 0.100
D. 0.150

Correct Option: Answer is A
Concentration (C) = number of moles (n) / volume (v)
i.e C = n/v
n = Cv
Concentration of H2SO4 = 0.01 mol dm-3
volume of H2SO4 = 20cm3 = 0.02 dm3
Thus, the number of moles of H2SO4 = 0.01 x 0.02 = 0.0002 mol
volume of NaOH solution = 20cm3 = 0.02 dm3
Therefore, concentration of NaOH solution = number of moles of H2SO4 / volume of NaOH solution
= 0.0002/0.02
0.01 = mol dm-3

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