Which of the following contains maximum number of atoms?

0.16 g C H_{3}

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0.28 g N_{2}

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0.36 g C_{6} H_{12} O_{6}

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1.1 g C O_{2}

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Solution

The correct option is D $1.1 g C O_{2}$
No. of atoms in $C H_{3}$ = $N_{a} \times m o l e$
= $N_{a} \times \frac{0.16}{15}$

No. of atoms in $N_{2}$ = $N_{a} \times m o l e$
= $N_{a} \times \frac{0.28}{28}$

No. of atoms in $C_{6} H_{12} O_{6} = N_{a} \times \frac{0.36}{180}$

No. of atoms in $C O_{2} = N_{a} \times \frac{1.1}{44}$
Hence $1.1$ g of $C O_{2}$ has the maximum no. of atoms.
Thus D is the correct option.

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