Number of electrons present in 3.6 mg of `NH_(4)^(+)` are : `(N_(A) = 6 xx 10^(23))`

Options

A. `1.2xx10^(21)`
B. `1.2xx10^(20)`
C. `1.2xx10^(22)`
D. `2xx10^(-3)`

Answer

A

Explanation

Number of electrons in one ion of `NH_(4)^(+)=10`
Number of ions in 3.6 mg of `NH_(4)^(+)`
`=(3.6xx10^(-3))/(18)xx6.022xx10^(23)=1.2xx10^(20)`
`:.` `NH_(4)^(+)`
`:.` Number of eletrons in 3.6 mg of `=1.2xx10^(20)xx10=1.2xx10^(21)`