Number of electrons present in 3.6 mg of `NH_(4)^(+)` are : `(N_(A) = 6 xx 10^(23))`
A
`1.2 xx 10^(21)`
B
`1.2 xx 10^(20)`
C
`1.2 xx 10^(22)`
D
`2 xx 10^(-3)`
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