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Volume of N(2) at 1 atm, 273 K required ...

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)`]

A

`16xx10^(-16)cm^(2)`

B

`0.35m^(2)//g`

C

`39m^(2)//g`

D

`22400cm^(2)`

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The correct Answer is:
B
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