The number of atoms in $4.25 g$ of $N H_{3}$ is approximately:

4 \times 10^{23}

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2 \times 10^{23}

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1 \times 10^{23}

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6 \times 10^{23}

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Solution

The correct option is D $6 \times 10^{23}$
$17 g of N H_{3} = 4 N_{A} atoms$
$4.25 g of N H_{3} = \frac{4 N_{A}}{17} \times 4.25 atoms$
$= N_{A} atoms = 6 \times 10^{23}$

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