Question:
$3.12 \mathrm{~g}$ of oxygen is adsorbed on $1.2 \mathrm{~g}$ of platinum metal. The value of oxygen adsorbed per gram of the adsorbent at 1 atm and $300 \mathrm{~K}$ in $\mathrm{L}$ is ________________ $\left[\mathrm{R}=0.0821 \mathrm{~L}\right.$ atm $\left.\mathrm{K}^{-1} \mathrm{~mol}^{-1}\right\rceil$
Solution:
(2)
Moles of $\mathrm{O}_{2}=\frac{3.12}{32}=0.0975$
volume of $\mathrm{O}_{2}=\frac{\mathrm{nRT}}{\mathrm{D}}=\frac{0.0975 \times 0.082 \times 300}{1}$
$=2.3985 \mathrm{~L} \simeq 2.4 \mathrm{~L}$
volume of $\mathrm{O}_{2}$ absorbed per $\mathrm{gm}$ of $\mathrm{pt}=\frac{2.4}{1.2}=2$
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