1.

The number of iodine atoms (N) present in 1 "cm"^(3) of its 0.1 M solution is

Answer»

`6.02 xx 10^(23)`
`6.02 xx 10^(2)`
`6.02 xx 10^(19)`
`1.204 xx 10^(20)`

Solution :1 cc of 0.1 M `I_(2)` SOL. `=0.1/1000=10^(-4)` mol
`N = n xx N_(A)`
`=10^(-4) xx 6.02 xx 10^(23)` molecules
`=6.02 xx 10^(19)` molecules
`=2 xx 6.02 xx 10^(19)` atoms `=1.204 xx 10^(20)`


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