The pressure of the gas was found to decrease from 720 to 480 mm. When 5 g of sample of activated charcoal was kept in a flask of one litre capacity maintained at 27° C. If the density of characoal is 1.25 gm / mL. The volume of gas adsorbed per gm of charcoal at 480 mm of Hg is-

Question

The pressure of the gas was found to decrease from 720 to 480 mm. When 5 g of sample of activated charcoal was kept in a flask of one litre capacity maintained at 27° C. If the density of characoal is 1.25 gm / mL. The volume of gas adsorbed per gm of charcoal at 480 mm of Hg is- (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

mole adsorbed =((720-480)/760) × (1/0.82 × 300) =0.0128 mole ∴ vol. at 480 mm Hg =(0.0128 × 0.082 × 300/(480)760) =0.498 ℓ=500 ml ∴ Volume/gm = (500/5) = 100 ml

Q. The pressure of the gas was found to decrease from 720 to 480mm. When 5g of sample of activated charcoal was kept in a flask of one litre capacity maintained at 27∘C. If the density of characoal is 1.25gm/mL. The volume of gas adsorbed per gm of charcoal at 480mm of Hg is-

mole adsorbed =760(720−480)​×0.82×3001​ =0.0128 mole ∴ vol. at 480mmHg=760480​0.0128×0.082×300​ =0.498ℓ=500ml ∴ Volume/gm=5500​=100ml

mole adsorbed $= \frac{( 720 - 480 )}{760} \times \frac{1}{0.82 \times 300}$
$= 0.0128$ mole
$∴$ vol. at $480 mm H g = \frac{0.0128 \times 0.082 \times 300}{\frac{480}{760}}$
$= 0.498 ℓ = 500 m l$
$∴$ Volume/ $g m = \frac{500}{5} = 100 m l$