Question

3.6 g of oxygen is adsorbed on 1.2 g of a metal powder. What volume of oxygen is adsorbed per gram of the adsorbent at 1 atm and 300 K?

• A. $0.19L{g}^{-1}$
• B. $1L{g}^{-1}$
• C. $2.31L{g}^{-1}$
• D. $3.14L{g}^{-1}$

Answer 1

The correct option is C $2.31L{g}^{-1}$

Mass of $O_2$ adsorbed per gram of adsorbent =$\frac{3.6}{1.2}$= 3
Number of moles of $O_2$ per gram of adsorbent = $\frac{3}{32}$
Volume of $O_2$ per gram of adsorbent =$\frac{3}{32}\times \frac{\left(0.0821\right)\times \left(300\right)}{1}$
=2.31 $L{g}^{-1}$
You can use the approximation that $R\approx \frac{1}{12}Latm/molK$ and you get the answer to be 2.34