3 . 12g of oxygen is adsorbed on 1 . 2g of platinum metal. The...

3 . 12g $of oxygen is adsorbed on$ 1 . 2g $of platinum metal. The volume of oxygen adsorbed per gram of the adsorbent at$ 1atm and 300K in L $is <u> </u></p><p>\left[R = 0 . 0821Latm{K}^{- 1}{mol}^{- 1}\right]$

Correct answer

Step-by-step solution

$V = \dfrac{\dfrac{3 . 12}{32} \times 0 . 0821 \times 300}{1} = 2 . 40L$

$∵ 1 . 2gm$ adsorbs $2 . 40L$

$∴ 1gm$ adsorbs $2L$

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