The number of atoms in `4.25 g NH_(3)` is approximately:

Options

A. `1xx10^(23)`
B. `2xx10^(23)`
C. `4xx10^(23)`
D. `6xx10^(23)`

Answer

D

Explanation

`:' 17 g NH_(3)` contains `6xx10^(23)` molecules of `NH_(3)`
`:. 4.25 g NH_(3)` contains`=(6xx10^(23))/17xx4.25`
`:.` No. of atoms`=(6xx10^(23)xx4.25)/17xx4=6xx10^(23)`.