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Standard X

Chemistry

Volume of Gases and Number of Moles

One gram of a...

Question

One gram of activates carbon has a surface area of $1000 m^{2} .$ Considering complete coverage as well as monomolecular adsorption, how much ammonia at $S T P$ would be adsorbed on the surface of $\frac{44}{7} g$ carbon if a radius of a ammonia molecules is $10^{- 8} c m .$

7.46 L

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0.33 L

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44.8 L

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23.5 L

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Solution

The correct option is A $7.46 L$
$1 g$ activated carbon has surface area of $100 m^{2}$
$\frac{44}{7} g$ activated carbon has surface area of $\frac{44}{7} \times 100 m^{2}$ at $S T P$ let $N$ no. of ammonia molecules are adsorbed.
$∴ N π r^{2} = \frac{44}{7} \times 1000$
or, $N = 2 \times 10^{23}$
$∴ volume of ammonia adsorbed = 7.46 L$

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One gram of activates carbon has a surface area of 1000 m2. Considering complete coverage as well as monomolecular adsorption, how much ammonia at STP would be adsorbed on the surface of 447 g carbon if a radius of a ammonia molecules is 10−8 cm.

The correct option is A 7.46 L1 g activated carbon has surface area of 100 m2 447 g activated carbon has surface area of 447×100 m2 at STP let N no. of ammonia molecules are adsorbed. ∴Nπr2=447×1000 or, N=2×1023 ∴volume of ammonia adsorbed=7.46 L

One gram of activates carbon has a surface area of $1000 m^{2} .$ Considering complete coverage as well as monomolecular adsorption, how much ammonia at $S T P$ would be adsorbed on the surface of $\frac{44}{7} g$ carbon if a radius of a ammonia molecules is $10^{- 8} c m .$