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 $\_\_\_\_$ .
$$
\left[\mathrm{R}=0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]
$$
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Solution
$$
\mathrm{V}=\frac{\frac{3.12}{32} \times 0.0821 \times 300}{1}=2.40 \mathrm{l}
$$
$\because 1.2 \mathrm{gm}$ adsorbs 2.40 l
$\therefore 1 \mathrm{gm}$ adsorbs 2 l
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