The number of atoms present in 4-25 grams of NH 3 is approximatelyA- 2 - 1023 NH 3B- 1 - 1023C- 1-5 - 1023D- 6-02 - 1023

The correct option is D 6.02 X 1023 4.25/17 × 4 × 6.023 × 10^23 = 6.023 × 10^23 ...

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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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