Question

☺ How many atoms of oxygen are present in 300 grams of calcium carbonate(CaCO3) ?

Answer 1

❀ HERE IS YOUR ANSWER ❀

1 mole of Calcium carbonate contains = 3 mole of oxygen atoms

100 g of Calcium carbonate = 3 X 6.022 X 10²³ oxygen atoms

300 g of Calcium carbonate = 300 X 3 X 6.022 X 10²³/ 100 = 54.205 X 10²³ oxygen atoms

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Answer 2

★ Hey mate ★

here is your answer !!

mole = mass / molecular mass

mass = 300 gram

molecular mass of calcium carbonate =
CaCO3 => 100 g

mole = mass / molecular mass

=> 300 / 100

3 mol

hence , numbers of molecules =>

3 × 6.22 × 10 ²³ molecules

now , total number of atom is

O3 = 3

total number of atoms = 3

hence , the total number of (atoms ) present is =>

3× 3 × 6.22 × 10 ²³

=>9 × 6.22 × 10 ²³
or

55 .98 × 10 ²³

hope it helps you dear !!

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