3.12 , g of oxygen is adsorbed on 1.2 , g of | vedclass

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(A) First,calculate the number of moles of $O_2$ adsorbed:$n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{3.12 \, g}{32 \, g/mol} = 0.0975 \, mol$.

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$2$

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(A) First,calculate the number of moles of $O_2$ adsorbed:$n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{3.12 \, g}{32 \, g/mol} = 0.0975 \, mol$.Next,calculate the volume of $O_2$ using the ideal gas equation $PV = nRT$:$V = \frac{nRT}{P} = \frac{0.0975 \, mol 	imes 0.0821 \, L \, atm \, K^{-1} \, mol^{-1} 	imes 300 \, K}{1 \, atm} = 2.40 \, L$.Since $1.2 \, g$ of platinum adsorbs $2.40 \, L$ of $O_2$,the volume adsorbed per gram is:$\frac{2.40 \, L}{1.2 \, g} = 2 \, L/g$.

$3.12 \, g$ of oxygen is adsorbed on $1.2 \, g$ of platinum metal. The volume of oxygen adsorbed per gram of the adsorbent at $1 \, atm$ and $300 \, K$ in $L$ is ..... .
$[R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}]$

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$3.12 \, g$ of oxygen is adsorbed on $1.2 \, g$ of platinum metal. The volume of oxygen adsorbed per gram of the adsorbent at $1 \, atm$ and $300 \, K$ in $L$ is ..... .$[R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}]$