Question

number of oxygen atoms present in 25 grams of CaCO3 (mol. wt. 100)

Answer 1

no . of moles of CaCO3 =
$\frac{25 \: g}{100 \: g}$
this is 0.25 moles of CaCO3.
Since one Mole of CaCO3 contains 3 moles of O, so 0.25 moles of CaCO3 contains 0.25×3 =0.75 moles of O.
no. of atoms of O = 0.75 x 6.022 x 10^23
= 4.5 x 10^23 atoms