The surface area of 1 , g of activated charco | vedclass

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Question

(A) Radius of $NH_3$ molecule,$r = 10^{-8} \, cm = 10^{-10} \, m$. Area occupied by one $NH_3$ molecule $= \pi r^2 = \frac{22}{7} 	imes (10^{-10})^2

Vedclass · Accepted Answer

$7.46$

Vedclass · Answer

(A) Radius of $NH_3$ molecule,$r = 10^{-8} \, cm = 10^{-10} \, m$. Area occupied by one $NH_3$ molecule $= \pi r^2 = \frac{22}{7} 	imes (10^{-10})^2 = \frac{22}{7} 	imes 10^{-20} \, m^2$. Total surface area of $\frac{44}{7} \, g$ charcoal $= \frac{44}{7} 	imes 1000 \, m^2$. Number of $NH_3$ molecules adsorbed $= \frac{	ext{Total Area}}{	ext{Area per molecule}} = \frac{\frac{44}{7} 	imes 1000}{\frac{22}{7} 	imes 10^{-20}} = 2 	imes 10^{23}$. Moles of $NH_3$ adsorbed $= \frac{2 	imes 10^{23}}{6 	imes 10^{23}} = \frac{1}{3} \, mol$. Volume of $NH_3$ at $STP = 	ext{moles} 	imes 22.4 \, L = \frac{1}{3} 	imes 22.4 = 7.46 \, L$.

The surface area of $1 \, g$ of activated charcoal is $1000 \, m^2$. Assuming complete coverage and unimolecular adsorption,if the radius of an ammonia molecule is $10^{-8} \, cm$,then the volume of ammonia adsorbed at $STP$ on $44/7 \, g$ of carbon will be ........ $L$. $(N_A = 6 \times 10^{23})$

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