1.

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.


Discussion

No Comment Found

Related InterviewSolutions