The volume of nitrogen gas adsorbed at STP to form a monolayer on a porous solid surface is 22.4 cm³ g^-1. If the area occupied by one nitrogen gas molecule is 16.2 Ų, then the surface area (in cm² g^-1) of the solid is close to:

Concept:

The surface area occupied = Total number of molecules combining for complete monolayer coverage × Area of 1 molecule

At STP,

1 mole gas = 22.4 L

22.4 L = 6.022 × 10²³ molecules

22.4 L = N_A

No. of molecules in 1 L = \(\frac{N_A}{22.4}\)

No. of molecules in V_m = \(\frac{N_A}{22.4} × V_m\)

where V_m = volume corresponding to total a complete monolayer coverage.

Explanation:

∵ The volume of N₂ at STP required to form a monolayer on a porous solid surface is = 22.4 cm³ g^-1

→ Area occupied by one molecule of nitrogen gas molecule = 16.2 Å²

= 16.2 × 10^-16 cm²

we know that (1 L = 10³ cm³)

∴ 22400 cm³ of N₂ at STP contains N_A molecules of N₂

∴ 22.4 cm³g^-1 of N₂ contains

= \(\frac{22.4 × N_A}{22400}\)

= 6.022 × 10²⁰ molecules of N₂

∴ Surface area occupied by 6.022 × 10²⁰ molecules of N₂,

= 6.022 × 10²⁰ × 16.2 × 10^-16 cm²

= 97.55 × 10⁴

= 9.755 × 10⁵ cm² g^-1

Conclusion:-
So, the option '2' is correct.