The number of atoms present in $4.25 g r a m s$ of $N H_{3}$ is approximately

1 X 10²³

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1.5 X 10²³

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2 X 10²³ NH3

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6.02 X 10²³

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Solution

The correct option is D
6.02 X 10²³

The number of moles present: $\frac{4.25 g}{17 g / m o l}$
Number of moles of atoms: $\frac{4.25}{17} \times 4 m o l$
Number of atoms: $\frac{4.25}{17} \times 4 \times 6.022 \times 10^{23} a t o m s$
= $6.022 \times 10^{23} a t o m s$

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