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 ?

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

Tardigrade Admin · Accepted Answer

Mass of oxygen adsorbed per gram of metal powder =(3.6/1.2) g =3 g = m B y ideal gas equation, P V=n R T ∴ V=( mRT / MP )=(3 × 273 × 0.0821/32 × 1 atm ) =2.1 Lg -1

Q. $3.6$ gram of oxygen is adsorbed on $1.2 g$ of metal powder. What volume of oxygen adsorbed per gram of the adsorbent at $1 a t m$ and $273 K$ ?

$0.19 L g^{- 1}$

$1 L g^{- 1}$

$2.1 L g^{- 1}$

None of these

Mass of oxygen adsorbed per gram of metal powder
$= \frac{3.6}{1.2} g = 3 g = m$ B
y ideal gas equation, $P V = n RT$
$∴ V = \frac{m RT}{MP} = \frac{3 \times 273 \times 0.0821}{32 \times 1 a t m}$
$= 2.1 L g^{- 1}$

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

0.19Lg−1

1Lg−1

2.1Lg−1