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Rupal Kumar Dey
Asked: 2022-11-10T08:57:27+05:30 In:

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)`.

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)`.
Parabola
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  1. Density of sites `=6.023xx10^(14)`
    Total sites `=6.023xx10^(14)xx1000`
    `=6.023xx10^(17)`
    `20%` sites are occupied.
    So sites occupied`=(20)/(100)xx6.023xx10^(17)`
    `=12.04xx10^(16)`
    `PV=nRT`
    or `n+(0.001xx2.46xx10^(-3))/(0.821xx300)=9.98xx10^(-9)`
    Molecules`=9.98xx10^(-9)xx6.023xx10^(23)`
    `=6.02xx10^(16)`
    Sites occupied per molecule of nitrogen is
    `(“Number of sites”)/(“Number of molecules”)=(12.04xx10^(16))/(6.02xx10^(16))=2`

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Seema Ram Gopal Narine
Asked: 2022-11-08T14:36:17+05:30 In:

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)`.

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)`.
Parabola
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  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`

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