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Calculate the surface area of a catalyst that adsorbs $10^{3} {cm}^{3} of N_{2}$ (reduced to STP) per gram in order to form the monolayer. The effective area occupied by N₂ molecule on the surface is $1 .62 \times 10^{- 15} {cm}^{2}$

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$2520 \times 10^{5} {cm}^{2}$

$4350 m^{2}$

$3720 m^{2}$

$435 \times 10^{5} {cm}^{2}$

answer is B, D.

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Detailed Solution

One molecule area $= 1.62 \times 10^{- 15}$

$22400 ⟶ 6 \times 10^{23} molecules; 10^{3} ⟶ ? \frac{6 \times 10^{26}}{22400} then \frac{6 \times 10^{26}}{22400} \times 1 .62 \times 10^{- 15} area occupied = 435 \times 10^{5} c m^{2}$

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