Question

$3.6$ 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 $273 \,K$ ?

• A. $0.19\, Lg ^{-1}$
• B. $1\, Lg ^{-1}$
• C. $2.1\, Lg ^{-1}$
• D. None of these

Answer 1

The Correct Option is C

Mass of oxygen adsorbed per gram of metal powder
$=\frac{3.6}{1.2} g =3\, g = m$ B
y ideal gas equation, $P V=n R T$
$\therefore V=\frac{ mRT }{ MP }=\frac{3 \times 273 \times 0.0821}{32 \times 1 atm }$
$=2.1\, Lg ^{-1}$

Answer 2

\(\text{Mass of }O_2​ \text{ per gram of adsorbent} =\frac{3.6​}{1.2}=3.0\)
\(\text{Number of moles of }O_2​ \text{ per gram of adsorbent} =\frac{3}{32}​\)
\(\text{Volume of }O_2​ \text{ per gram of adsorbent}=\frac{3}{32}​\times{\frac{0.0821\times273}{1}}​=2.10 L\)

Answer 3

\(\text{The correct option is C: 2.31 }Lg^{−1}\)
\(\text{Mass of }O_2\text{ adsorbed per gram of adsorbent} =\frac{3.6}{1.2}= 3\)
\(\text{Number of moles of }O_2\text{ per gram of adsorbent} = \frac{3}{32}\)
\(\text{Volume of }O_2\text{ per gram of adsorbent} =\frac{3}{32}\times\frac{(0.0821)\times(300)}{1}\)
=2.31 Lg⁻¹
For the same approximation can be used i.e., R≈112\(R≈\frac{1}{12}\) L atm/mol K and you get the answer to be 2.34.