The number of atoms present in $10$ gms of $C a C O_{3}$ is:

5 N_{A}

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0.5 N_{A}

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5

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N_{A}

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Solution

The correct option is B $0.5 N_{A}$
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 $0.5 N_{A}$ atoms. Here, $N_{A}$ is the Avogadro number.

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