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

1 \times 10^{23}

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1.5 \times 10^{23}

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2 \times 10^{23}

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6 \times 10^{23}

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Solution

The correct option is C $6 \times 10^{23}$
Moles of $N H_{3} = \frac{4.25}{17} = 0.25$

Number of atoms in 0.25 moles of $N H_{3}$ is $0.25 \times 6.023 \times 10^{23} \times 4 = 6.023 \times 10^{23}$

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Similar questions

The number of atoms present in 4.25 g of $({NH}_{3})$ is approximately.

a) $1 \times 10^{23}$

b) $1.5 \times 10^{23}$

c) $0.25 \times 10^{23}$

d) $6.02 \times 10^{23}$

N H_{3}

N H_{3}

4.25 g r a m s

N H_{3}

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