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Number of electrons present in 3.6 mg of...

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

A

`1.2xx10^(21)`

B

`1.2xx10^(20)`

C

`1.2xx10^(22)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
Mole of `NH_(4)^(+)=(3.6xx10^(-3))/(18)=0.2xx10^(-3)=2xx10^(-4)`
no. of `NH_(4)^(+)` ions `= 2xx10^(-4)xx6.02xx10^(23)`
`=12.04xx10^(+19)=1.2xx10^(20)`
no. of electrons in one `NH_(4)^(+)` ions = 10
`therefore` Total no. of electrons `= 10xx1.2xx10^(20)`
`=1.2xx10^(21)`
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