Question

A sample of charcoal weighing 6.0 g was brought into contact with a gas contained in a vessel of 1 litre capacity at 27°C. The pressure of the gas falls from 780 to 400 torr. The density of charcoal sample was 1.5 g/mL. The volume of gas (at STP) that is adsorbed per g (mL/g) of the adsorbent under the condition of experiment is (Round off upto two decimal places.)
(Use R = $\frac{1}{12}Latm/mol$)

Answer 1

We know, PV=nRT
So, let $n_1=\frac{\frac{780}{760}\times 1}{R\times 300}$

$n_2=\frac{\frac{400}{760}\times 1}{R\times 300}$

$\therefore$ $n_1-n_2=\frac{\frac{380}{760}\times 1}{\frac{1}{12}\times 300}=0.02mol$,
So, volume of the gas adsorbed at STP = $(0.02\times 22400)$ mL = 448 mL
Weight of the adsorbent = 6 g
$\therefore$ volume of the gas adsorbed per g of adsorbent = $\frac{448}{6}$ = 74.67 mL/g.