$3.12 g$ of oxygen is adsorbed on $1.2 g$ of platinum metal. The volume of oxygen adsorbed per gram of the adsorbent at $1 a t m$ and $300 K$ in $L$ is _____.

[ $R = 0.0821 L a t m K^{- 1} m o l^{- 1}$ ]

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Solution

$P = 1 a t m T = 300 K Mass of oxygen adsorbed on 1 g of platinum = \frac{3.12}{1.2} = 2.6$
$No. of moles of oxygen adsorbed on 1 g of platinum = \frac{2.6}{32} = 0.0812$
Using ideal gas equation,
$P V = n R T$
$V = \frac{0.0812 \times 0.0821 \times 300}{1} = 1.99 L \approx 2 L$

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$3.12 g$ of oxygen is adsorbed on $1.2 g$ of platinum metal. The value of oxygen adsorbed per gram of the adsorbent at $1 a t m$ and $300 K$ in $L$ is _______.
$[R = 0.0821 L a t m K^{- 1} m o l^{- 1}]$

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1 a t m

and

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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−1mol−1]

P=1 atmT=300 KMass of oxygen adsorbed on 1 g of platinum=3.121.2=2.6 No. of moles of oxygen adsorbed on 1 g of platinum=2.632=0.0812 Using ideal gas equation, PV=nRT V=0.0812×0.0821×3001=1.99 L≈2 L

$3.12 g$ of oxygen is adsorbed on $1.2 g$ of platinum metal. The volume of oxygen adsorbed per gram of the adsorbent at $1 a t m$ and $300 K$ in $L$ is _____.

[ $R = 0.0821 L a t m K^{- 1} m o l^{- 1}$ ]

$P = 1 a t m T = 300 K Mass of oxygen adsorbed on 1 g of platinum = \frac{3.12}{1.2} = 2.6$
$No. of moles of oxygen adsorbed on 1 g of platinum = \frac{2.6}{32} = 0.0812$
Using ideal gas equation,
$P V = n R T$
$V = \frac{0.0812 \times 0.0821 \times 300}{1} = 1.99 L \approx 2 L$