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The number of Hydrogen ions in 1.0 $dm^{3}$ of 0.02\(moldm^...

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.

$2.4 \times 10^{23}$

View answer & explanation

Correct answer is A

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

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

= $0.1204 \times 10^{23} = 1.204 \times 10^{22}$

$\approxeq 1.2 \times 10^{22}$

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