One gram of activated carbon has a surface area of 1000 m². Considering complete coverage as well as monomolecular adsorption, how much ammonia at 1 atm and 273 K would be adsorbed on the surface of 44/7g carbon if radius of a arnmonia molecules is 10^-8 cm.
[Given: N_A - 6 x l0²³mol^-1]

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0.33L

7.46L

44.8 L

23.5 L

answer is A.

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Detailed Solution

$Total surface area of carbon = \frac{44}{7} \times 10^{7} {cm}^{2}$
$r = 10^{- 8} cm$
$Surface area of {NH}_{3} = {πr}^{2}$
$= \frac{22}{7} \times 10^{- 16} {cm}^{2}$
$No. of {NH}_{3} molecules adsorbed = \frac{\frac{44}{7} \times 10^{7}}{\frac{22}{7} \times 10^{- 16}}$
Vol. of NH₃ adsorbed at I atm 273 K
$= \frac{2 \times 10^{23}}{6 \times 10^{23}} \times 22 .4 = 7 .46 L$