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Calculate no. of carbon and oxygen atoms present in 11.2 litres of `CO_(2)` at N.T.P |
Answer» Step I : No. of `CO_(2)` molecules in 11.2 litres `22.4` litres of `CO_(2)` at N.T.P = 1 gram mol 11.2 litres of `CO_(2)` at N.T.P `= ((1 "gram mol"))/((22.4 "litres"))xx(11.2"litres")=0.5` gram mol Now 1 gram mole of `CO_(2)` contain molecules `= 6.022 xx 10^(23)` `:. 0.5` gram mole of `CO_(2)` contain molecules `= 6.022xx10^(23)xx0.5=3.011xx10^(23)` Step II. No. of carbon and oxygen atoms in `3.011 xx 10^(23)` molecules of `CO_(2)` 1 molecule of `CO_(2)` contains carbon atoms = 1 `:. 3.011 xx 10^(23)` Similarly, 1 molecule of `CO_(2)` contains oxygen atoms = 2 `:. 3.011 xx 10^(23)` molecules of `CO_(2)` will contain oxygen atoms `= 2 xx 3.011 xx 10^(23) = 6.022 xx 10^(23)` atoms. |
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