Volume of `N_(2)` at 1 atm, 273 K required to form a monolayer on the surface of iron catalyst is `8.15ml//gm` of the adsorbent. What will be the surface area of the adsorbent per gram if each nitrogen molecule occupies `16 xx 10^(-22)m^(2)`?
[Take : `N_(A)=6xx10^(23)`]

Options

A. `16xx10^(-16) cm^(2)`
B. `0.35 m^(2)//g`
C. `39 m^(2)//g`
D. `22400 cm^(2)`

Answer

B

Explanation

The volume of `N_(2)` at STP required to cover the ion surface with monolayer `=8.15 ml gm^(-1)`
Area occupied by single molecule `=16xx10^(-18) cm^(2)`
`22400` ml of `N_(2)` at STP contains `=N_(A)` molecule of `N_(2)`
`:. 8.15 ……………. =(8.15xxN_(A))/(22400)=2.19xx10^(20)` molecule of `N_(2)`
Area occupied by `2.19xx10^(20)` molecule of `N_(2)=2.19xx10^(20)xx16xx10^(-18) cm^(2)=35.06xx10^(2) cm^(2)`
surface area of the iron adsorbed `=0.35 m^(2) gm^(-1)`
In short `A=("Volume covered by the" N_(2) "molecule"xxN_(A)xx"Area occupied by single molecule")/22400`