Which of the following samples will have same number of atoms as $6.4$ g of $C H_{4}$ ?

8.5 g

N H_{3}

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12 g

H_{2} O

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32 g

S O_{2}

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22 g

C O_{2}

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Solution

The correct options are
A $8.5 g$ of $N H_{3}$
B $12 g$ of $H_{2} O$
Number atoms of $6.4$ gm of ${C H}_{4} = \frac{5 \times 6.4}{16} \times 6.023 \times 10^{23}$
$= 2.0 \times 6.023 \times 10^{23}$
A) Number atoms of ${N H}_{3} = 4 \times \frac{8.5}{17} \times 6.023 \times 10^{23}$
$= 2 \times 6.023 \times 10^{23}$
B) Number atoms of $12$ gm of $H_{2} O = \frac{3 \times 12}{18} \times 6.023 \times 10^{23}$
$= 2 \times 6.023 \times 10^{23}$
C) Number of atoms of $32$ gm of ${S O}_{2} = \frac{3 \times 32}{64} \times 6.023 \times 10^{23}$
$= 1.5 \times 6.023 \times 10^{23}$
D) Number of atoms of $22$ gm of ${C O}_{2} = \frac{3 \times 22}{44} \times 6.023 \times 10^{23}$
$= 1.5 \times 6.023 \times 10^{23}$
$∴$ Correct answer are $A$ and $B$ .

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