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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)`. |
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Answer» 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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