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