How many atoms of each constituent element are present in 50 g of `CaCO_(3)` ?

Answer

(i) `Ca = 3.01 xx 10^(23)` (ii) `C = 3.01 xx 10^(23)` (iii) `O = 9.03 xx 10^(23)`

Explanation

Gram molecular mass of `CaCO_(3) = 40+12 +3xx16 =100.0g`
100.0 grams of `CaCO_(3)=1` gram mol
50.0 grams of `CaCO_(3) = (50.0)/(100.0) =0.5` gram mol.
One gram mole of `CaCO_(3)` contain Ca atoms `= 6.022 xx 10^(23)` atoms
0.5 gram mole of `CaCO_(3)` contain Ca atoms `= 6.022xx10^(23) xx 0.5 = 3.01 xx 10^(23)`
One gram mole of `CaCO_(3)` contain carbon atoms `= 6.022 xx 10^(23)` atoms
0.5 gram mole `CaCO_(3)` contain carbon atoms `= 3.01 xx 10^(23)` atoms
One gram mole of `CaCO_(3)` contain oxygen atoms `= 3 xx 6.022 xx 10^(23)` atoms
0.5 gram mole of `CaCO_(3)` contain oxygen atoms `= 0.5 xx 3 xx 6.022 xx 10^(23) = 9.03 xx 10^(23)` atoms.