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I cc `N_(2)O` at NTP containsA. `(1.8)/(224)xx10^(22)` atomsB. `(6.02)/(22400)xx10^(23)` moleculesC. `(1.32)/(224)xx10^(23)` electronsD. All of the above |
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Answer» Correct Answer - D At NTP 22400 c c of `N_(2)O` contains `=6.02xx10^(23)` molecules `:. 1 c c N_(2)O` will contain `=(6.02xx10^(23))/(22400)` molecules In `N_(2)O` molecules , number of atoms =2+1=3 Thus , number of atoms `=(3xx6.02xx10^(23))/(22400)` atoms `=(1.8xx10^(22))/(224)` atoms In `N_(2)O` molecule , number of electrons `=7+7+8=22` Hence , number of electrons `=(6.02xx10^(23))/(22400)xx22` electrons `=(1.32xx10^(23))/(224)` electrons |
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