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

Options

A. `6.023xx10^(23)`
B. `4xx6.023xx10^(23)`
C. `1.7xx10^(24)`
D. `4.25xx6.023xx10^(23)`

Answer

A

Explanation

Number of atoms in 4.25g of `NH_(3)=`number of moles of `NH_(3)xx`Avogadro's number`xx`number of atoms per molecule of `NH_(3)`
`=(4.25)/(17)xx6.023xx10^(23)xx4`
`=6.023xx10^(23)` atoms
`=N_(A)=6.023xx10^(23)` atoms