Question

Around `20%` surface sites have adsorbed `N_(2)`. On heating `N_(2)` gas evolved form sites and were collected at 0.001 atm and 298 K in a container of volume `2.46cm^(3)` the density of surface sites is `6.023xx10^(14)cm^(-2)` and surface area is `1000cm^(2)` find out the number of surface sites occupied per molecule of `N_(2)`.

Answer 1

Correct Answer - -2
Partial ideal gas law : `pV = nRT`
`n(N_(2)) = (pV)/(RT) = (0.001 xx 2.46)/(0.082 xx 298) = 10^(7)`
`rArr` Number of molecules of `N_(2) = 6.023 xx 10^(23) xx 10^(-7)`
`= 6.023 xx 10^(16)`
Now, total surface sites available
`= 6.023 xx 10^(14) xx 1000 = 6.023 xx 10^(17)`
Surface sites used in adsorption `= (20)/(100) xx 6.023 xx 10^(17)`
`= 2 xx 6.023 xx 10^(16)`
`rArr` Sites occupied per molecules
`= ("Number of sites")/("Number of molecules") = (2 xx 6.023 xx 10^(16))/(6.023 xx 10^(16)) = 2`