The number of Hydrogen ions in 1.0\(dm^{3}\) of 0.02\(moldm^{-3}\) tetraoxosulphate(VI) acid is \([N_{A} = 6.02 \times 10^{23}]\)
-
\(1.2 \times 10^{22}\)
- \(1.2 \times 10^{23}\)
- \(2.4 \times 10^{22}\)
- \(2.4 \times 10^{23}\)
The correct answer is: A
Explanation
\(1mole = 6.02 \times 10^{23} ions\)
\(0.02moldm^{-3} = 0.02 \times 6.02 \times 10^{23}\)
=\(0.1204 \times 10^{23} = 1.204 \times 10^{22}\)
\(\approxeq 1.2 \times 10^{22}\)