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