2.0 , g of H_2 gas is adsorbed on 2.5 , g of | vedclass

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(C) First,calculate the number of moles of $H_2$ gas: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{2.0 \, g}{2.0 \, g/mol} = 1.0 \, mol$.Using t

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$9960$

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(C) First,calculate the number of moles of $H_2$ gas: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{2.0 \, g}{2.0 \, g/mol} = 1.0 \, mol$.Using the ideal gas equation $PV = nRT$,we find the volume $V$ of the gas:$V = \frac{nRT}{P} = \frac{1.0 \, mol 	imes 0.083 \, L \, bar \, K^{-1} \, mol^{-1} 	imes 300 \, K}{1 \, bar} = 24.9 \, L$.Convert the volume to $mL$: $24.9 \, L = 24900 \, mL$.The volume of gas adsorbed per gram of adsorbent is calculated as:$	ext{Volume per gram} = \frac{	ext{Total volume}}{	ext{Mass of adsorbent}} = \frac{24900 \, mL}{2.5 \, g} = 9960 \, mL/g$.

$2.0 \, g$ of $H_2$ gas is adsorbed on $2.5 \, g$ of platinum powder at $300 \, K$ and $1 \, bar$ pressure. The volume of the gas adsorbed per gram of the adsorbent is $..... \, mL$.
(Given: $R = 0.083 \, L \, bar \, K^{-1} \, mol^{-1}$)

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$2.0 \, g$ of $H_2$ gas is adsorbed on $2.5 \, g$ of platinum powder at $300 \, K$ and $1 \, bar$ pressure. The volume of the gas adsorbed per gram of the adsorbent is $..... \, mL$.(Given: $R = 0.083 \, L \, bar \, K^{-1} \, mol^{-1}$)