The number of Hydrogen ions in 1.0\(dm^{3}\) of 0.02\(moldm^{-3}\) tetraoxosulphate(VI) acid is   \([N_{A} = 6.02 \times 10^{23}]\)

A.

\(1.2 \times 10^{22}\)

B.

\(1.2 \times 10^{23}\)

C.

\(2.4 \times 10^{22}\)

D.

\(2.4 \times 10^{23}\)

Correct answer is A

\(1mole = 6.02 \times 10^{23} ions\)

\(0.02moldm^{-3} = 0.02 \times 6.02 \times 10^{23}\)

=\(0.1204 \times 10^{23} = 1.204 \times 10^{22}\)

\(\approxeq 1.2 \times 10^{22}\)