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

Options

A. `6.02 xx 10^(23)`
B. `6.02 xx 10^(2)`
C. `6.02 xx 10^(19)`
D. `1.204 xx 10^(20)`

Answer

D

Explanation

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)`