1.

One gram of activated carbon has a surface area of 1000cm^2 considering complete coverage as well as monomolecular adsorption. How much ammonia at latm and 273K would be adsorbed on the surface of 44/7gr. carbon. If radius of a ammonia molecule is 10^(-8)cm .........

Answer»


Solution :Total surface area of carbon `44/7 xx 10^(7) cm^(2) , r = 10^(-8) cm`
surface area of `NH_3 = pi r^2 = 22/7 xx 10^(-16) cm^2`
No. of `NH_3` molecules adsorbed ` = ((44)/(7) xx 10^(7))/(22/7 xx 10^(-16)) = 2 xx 10^(23)`
vol. of `NH_3` adsorbed at STP `= (2 xx 10^(23))/(6 xx 10^(23)) xx 22.4 = 7.46 ` lit


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