Question

I cc `N_(2)O` at NTP contains
A. `(1.8)/(224)xx10^(22)` atoms
B. `(6.02)/(22400)xx10^(23)` molecules
C. `(1.32)/(224)xx10^(23)` electrons
D. All of the above

Answer 1

Correct Answer - D
At NTP 22400 c c of `N_(2)O` contains `=6.02xx10^(23)` molecules
`:. 1 c c N_(2)O` will contain `=(6.02xx10^(23))/(22400)` molecules
In `N_(2)O` molecules , number of atoms =2+1=3
Thus , number of atoms `=(3xx6.02xx10^(23))/(22400)` atoms
`=(1.8xx10^(22))/(224)` atoms
In `N_(2)O` molecule , number of electrons
`=7+7+8=22`
Hence , number of electrons `=(6.02xx10^(23))/(22400)xx22` electrons
`=(1.32xx10^(23))/(224)` electrons