An oxide of N has vapour density `46`. Find the total number of electron in its `92 g.(N_(A) = ` Avogdro's number )

Explanation

Let the formule of nitrogen oxide be `N_(2)O_(x)`.
Molecular mass of nitrogen oxide `= 2xx14+16x`
Molecular mass from vapour density `=xxV.D.=2xx46=92`.
On comparing , `28+16x=92 or x=(92-28)/(16)=4`
`:. " " ` Formula of nitrogen oxide `=N_(2)O_(4)`
No. of electrons in one molecule of `N_(2)O_(4)=2xx7+4xx8=46`
No. of electrons in Avogadro's molecules (92g) of `N_(2)O_(4)=46xx6.022xx10^(23)=2.77xx10^(25)` electron