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

Option: 1

9960


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

\mathrm{Given, R=0.083 \mathrm{~L}\, bar \, \mathrm{K}^{-1} \, mol^{-1}}

\mathrm{mass \: of\: \mathrm{H}_{2} \: adsorbed =2 \mathrm{~g}.}
\mathrm{mass \: of \: platinum \: powder=2.5 \mathrm{~g}.}

\mathrm{T=300 \mathrm{~K}}
\mathrm{p=1 bar}

\mathrm{Volume\: of \: \mathrm{H}_{2}=\frac{n R T}{P}=\left(\frac{2}{2} \times \frac{0.083 \times 300}{1}\right) \mathrm{L}}
                                                \mathrm{=24.9 \, \mathrm{L}}
                                                \mathrm{=24900 \, \mathrm{mL}}

\mathrm{2.5 \mathrm{~g} \: \text{of Platinum absorbs}\;= 24900\: \mathrm{ml} \: of \: \mathrm{H}_{2}}
\mathrm{So,1 \mathrm{~g} \: \text{of Platinum absorbs}\;= \frac{ 24900}{2.5}\: \mathrm{ml} \: of \: \mathrm{H}_{2}}
                                                             \mathrm{=9960\, ml\: of\: H_{2}}
9960 is the correct answer.

Posted by

Sumit Saini

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