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: Option 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})