`10^(21)` molecules are removed from 200 mg of `CO_(2)`.The moles of `CO_(2)` left are:

Options

A. `2.88xx10^(-3)`
B. `28.8xx10^(-3)`
C. `288xx10^(-3)`
D. `28.8xx10^(3)`

Answer

A

Explanation

`6.02 xx 10^(23)` molecules = 1 mol
`1 xx 10^(21)` molecule ` = (1xx 10^(21))/(6.02xx 10^(23))` mol
`= 1.66 xx 10^(-3)` mol
`200 " mg "CO_(2) = (200xx 10^(-3))/44` mol
` = 4.54 xx 10^(-3)` mol
Thus,
`CO_(2) " left" = (4.54-1.66)xx 10^(-3)`
` =2. 88 xx 10^(-3) ` mol