Question

what is the no. of atoms in 22 g of CaCo3​

Answer 1

Answer:

The molecular weight of calcium carbonate is 100 g.

10 g of calcium carbonate corresponds to 0.1 mole.

One molecule of calcuim carbonate contains 5 atoms. Hence, 0.1 mole of calcium carbonate will contain 0.5Na atoms. Here, Na is the Avogadro number.

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Answer 2

Explanation:

Atomic mass: Ca=40,C=12,O=16

1mole ofCaCO3=100g. (40×1+12+16×3=100)

1mole of CaCO3 =6.022×10^23 molecules

Then,100g CaCO3 =6.022×10^23 molecules

For 22g CaCO3

=(6.022×10^23×22)÷100 molecules.

=(6.022×10^23×22×5)÷ 100 atoms.

( CaCO3 contains 5 atoms)