Question

1.2046 x 10^23 molecules of calcium carbonate​

Answer 1

Molar mass of CaCO3 = 40+12+3×16 = 100 g

✨No. of moles of CaCO3

➡ No. of molecules/Avogadro constant

➡ 6.022 × 1023/ 6.022 × 1023

➡1 mole

Mass of CaCO3

=➡No. of moles × molar mass

➡ 1 × 100 g = 100 g.

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Answer 2

Answer:

Molar mass of CaCO3 = 40+12+3×16 = 100 g

4G

45

No. of moles of CaCO3

➡No. of molecules/Avogadro constant

➡6.022 × 1023/ 6.022 × 1023

➡1 mole

Mass of CaCO3

=>No. of moles x molar mass

➡1 x 100 g = 100 g.

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