2 . 0g of H _ 2 gas is adsorbed on 2 . 5g of platinum powder at...

$2 . 0g$ of ${H}_{2}gas$ is adsorbed on $2 . 5g$ of platinum powder at $300K$ and $1$ bar pressure. The volume of the gas adsorbed per gram of the adsorbent is $mL$

(Given : $R = 0 . 083L$ bar ${K}^{- 1}{mol}^{- 1}$ )

Correct answer

9960

Step-by-step solution

{V}_{{H}_{2}} = \dfrac{nRT}{P}{V}_{{H}_{2}}\left(g\right) = \left(\dfrac{2}{2}\right)\dfrac{\left(0 . 083\right)300}{1} $</p><p>$ ∴ $Volume of$ {H}_{2}\left(g\right) $adsorbed on$ 2 . 5Pt = 24 . 9L $</p><p>$ ∴ $Volume of$ {H}_{2} $\left(g\right)$ adsorbed on $1gofPt = \dfrac{24 . 9}{2 . 5} = 9 . 96L = 9960ml$

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