2.0 g of H 2 gas is adsorbed on 2.5 g of platinum powder at 300 K and 1 bar pressure. The volume of the gas adsorbed per gram of the adsorbent is mL. (Given: R =0.083 L bar K -1 mol -1 )

Question

2.0 g of H 2 gas is adsorbed on 2.5 g of platinum powder at 300 K and 1 bar pressure. The volume of the gas adsorbed per gram of the adsorbent is mL. (Given: R =0.083 L bar K -1 mol -1 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Volume of H 2=( nRT / p ) =(2/2) × (0.083 × 300/1) =24.92 L =24900 mL So 1 g platinum adsorb =(24900/2.5) mLH 2 =9960

Q. $2.0 g$ of $H_{2}$ gas is adsorbed on $2.5 g$ of platinum powder at $300 K$ and $1$ bar pressure. The volume of the gas adsorbed per gram of the adsorbent is _______ $m L$ .
(Given : $R = 0.083 L$ bar $K^{- 1} m o l^{- 1}$ )

Volume of $H_{2} = \frac{n RT}{p}$
$= \frac{2}{2} \times \frac{0.083 \times 300}{1}$
$= 24.92 L$
$= 24900 m L$
So $1 g$ platinum adsorb $= \frac{24900}{2.5} m L H_{2}$
$= 9960$

Q. 2.0g of H2​ gas is adsorbed on 2.5g of platinum powder at 300K and 1 bar pressure. The volume of the gas adsorbed per gram of the adsorbent is _______mL. (Given : R=0.083L bar K−1mol−1 )

Volume of H2​=pnRT​ =22​×10.083×300​ =24.92L =24900mL So 1g platinum adsorb =2.524900​mLH2​ =9960

Q. $2.0 g$ of $H_{2}$ gas is adsorbed on $2.5 g$ of platinum powder at $300 K$ and $1$ bar pressure. The volume of the gas adsorbed per gram of the adsorbent is _______ $m L$ .
(Given : $R = 0.083 L$ bar $K^{- 1} m o l^{- 1}$ )

Volume of $H_{2} = \frac{n RT}{p}$
$= \frac{2}{2} \times \frac{0.083 \times 300}{1}$
$= 24.92 L$
$= 24900 m L$
So $1 g$ platinum adsorb $= \frac{24900}{2.5} m L H_{2}$
$= 9960$