3.6 gram of oxygen is adsorbed on $$1.2\,g$$  of metal powder. What volume of oxygen adsorbed per gram of the adsorbent at $$1\,atm$$  and $$273\,K?$$

Solution :
Mass of $${O_2}$$ per gram of adsorbent $$ = \frac{{3.6}}{{1.2}} = 3$$
No. of moles of $${O_2}$$ per gram of adsorbent $$ = \frac{3}{{32}}$$
Volume of $${O_2}$$ per gram of adsorbent $$ = \frac{3}{{32}} \times \frac{{0.0821 \times 273}}{1} = 2.10$$