Question

The number of oxygen molecules present in 100 grams of limestone

Answer 1

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Molecular mass is CaCO3 = 40+12+(3×16) = 100g [Molecular weight of Ca=40, C=12, O =16]

1 mole of oxygen atoms = 6.022 × 10^23 oxygen atoms [ Avogadro's Number]

1 mole of calcium carbonate = 100g of calcium carbonate

100g of calcium carbonate(100,000mg of calcium carbonate

100g of calcium carbonate(100,000mg of calcium carbonate) has 3 moles of oxygen atoms = 3× 6.022×10^23 oxygen atoms

Therefore, 100mg of calcium carbonate has (3×6.022×10^23×100)/100,000 oxygen atoms

=> 100mg of calcium carbonate has (3×6.022×10^23)/1000 oxygen atoms

=> 100 mg of calcium carbonate has 3×6.022×10^20 oxygen atoms

Answer 2

Answer:1.5N

Explanation:48(wt of 03)/32(wt of o2)×N(avagadro.no)

=3/2N=1.5N

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