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(A) Volume of $N_2$ at $STP = 1.30 \, cm^3 \, g^{-1} = 1.30 	imes 10^{-3} \, L \, g^{-1}$.Moles of $N_2$ adsorbed $= \frac{1.30 	imes 10^{-3} \, L}{

Vedclass · Accepted Answer

$5.57$

Vedclass · Answer

(A) Volume of $N_2$ at $STP = 1.30 \, cm^3 \, g^{-1} = 1.30 	imes 10^{-3} \, L \, g^{-1}$.Moles of $N_2$ adsorbed $= \frac{1.30 	imes 10^{-3} \, L}{22.4 \, L \, mol^{-1}} = 5.8036 	imes 10^{-5} \, mol \, g^{-1}$.Number of $N_2$ molecules $= (5.8036 	imes 10^{-5}) 	imes (6.022 	imes 10^{23}) \approx 3.495 	imes 10^{19} \, 	ext{molecules} \, g^{-1}$.Area occupied by one molecule $= 0.16 \, nm^2 = 0.16 	imes 10^{-18} \, m^2$.Total surface area $= (3.495 	imes 10^{19}) 	imes (0.16 	imes 10^{-18}) \, m^2 \, g^{-1} = 5.592 \, m^2 \, g^{-1}$.Rounding to the nearest value provided,the answer is $5.57 \, m^2 \, g^{-1}$.

Assuming Langmuir monolayer adsorption,the volume of nitrogen gas adsorbed on a sample of silica gel at $STP$ is $1.30 \, cm^3 \, g^{-1}$. The area occupied by a nitrogen molecule is $0.16 \, nm^2$. The surface area of silica gel per gram is ......... $m^2 \, g^{-1}$.

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