The total number of atoms present in `10.6g` of `Na_(2) CO_(3)` will be

Options

A. `1.89xx10^(23)`
B. `3.61xx10^(23)`
C. `24.1xx10^(23)`
D. `12.0xx10^(23)`

Answer

B

Explanation

`10.6g Na_(2) CO_(3) xx (1 "mol" Na_(2) CO_(3))/(106 g Na_(2) CO_(3))`
`xx (6.022xx10^(23) "formula units")/(1 mol Na_(2) CO_(3)) xx (6 "atoms")/(1 "formula unit")`
`= 3.61xx10^(23)`