Which of the following compounds contains maximum number of atoms?

18gm of

H_{2} O

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18gm of

O_{2}

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18gm of

C O_{2}

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18gm of

C H_{4}

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Solution

The correct option is D 18gm of $C H_{4}$

$N o . o f a t o m s = \frac{M a s s \times a t o m i c i t y}{M o l a r M a s s} \times N_{A}$
Using the above formula:
a) In 18gm of $H_{2} O$ :
No. of atoms= $\frac{18 \times 3}{18} \times N_{A} = 3 N_{A}$

b) In 18gm of $O_{2}$ :
No of atoms= $\frac{18 \times 2}{32} \times N_{A} = 1.12 N_{A}$

c) In 18gm of $C O_{2}$ :
No. of atoms= $\frac{18 \times 3}{44} \times N_{A} = 1.23 N_{A}$

d) In 18gm of $C H_{4}$ :
No. of atoms= $\frac{18 \times 5}{16} \times N_{A} = 5.63 N_{A}$
Therefore, 18gm $C H_{4}$ has a maximum number of atoms.

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