I cc `N_(2)O` at NTP contains

Options

A. `(1.8)/(224) xx 10^(22)` atoms
B. `(6.02)/(22400) xx 10^(23)` molecules
C. `(1.32)/(224) xx 10^(23)` electrons
D. All of the above

Answer

D

Explanation

At NTP 22400 cc of `N_(2)O=6.02 xx 10^(23)` molecules
`:. "1 cc" N_(2)O=(6.02 xx 10^(23))/(22400)` molecules
`= (3 xx 6.02 x 10^(23))/(22400) "atoms"
=(1.8)/(224) xx 10^(22)` atoms
No. of electrons in a molecule of `N_(2)O = 7+7+8 = 22`
Hence no. of electrons
`=(6.02 xx 10^(23))/(22400) xx 22` electrons `=(1.32 xx 10^(23))/(224)`