The number of atoms of oxygen present in 10.6 g of `Na_2 CO_3` will be

Options

A. `6.02 xx 10^(22)`
B. `12.04 xx 10^(22)`
C. `1.806 xx 10^(23)`
D. `31.8`

Answer

C

Explanation

1 mole i.e. 106 g `Na_2 CO_2` contains
` = 3 xx 6.023 xx 10^(23) = 18.06 xx 10^(23)` oxygen atoms
`=( 18.06 xx 10^(23) xx 10.6 )/( 106 ) = 1.806 xx 10^(23)`