The number of atoms present in 4-25 grams of NH 3 is approximately-A- 6-02 - 1023B- 1-5 - 1023C- 24-08 - 1023D- 0-5 - 1023

The correct option is A 6.02×1023The number of moles present: 4.25 g/17 g/mol Number of moles of atoms: 4.25/17×4 mol Number of atoms:4.25/17× 4× 6.022×10^23 a ...

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Standard VII

Chemistry

Sub-Atomic Particles

The number of...

Question

The number of atoms present in 4.25 grams of NH3 is approximately:

6.02 \times 10^{23}

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1.5 \times 10^{23}

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24.08 \times 10^{23}

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0.5 \times 10^{23}

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Solution

The correct option is A $6.02 \times 10^{23}$
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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Similar questions

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

Q. The number of atoms present in

4.25 g r a m s

N H_{3}

is approximately

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

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) $6.02 \times 10^{23}$

Q. The number of atoms in 4.25 g of

N H_{3}

is approximately

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The number of atoms present in 4.25 grams of NH3 is approximately:

The correct option is A 6.02×1023The number of moles present: 4.25 g17 g/mol Number of moles of atoms: 4.2517×4 mol Number of atoms:4.2517×4×6.022×1023 atoms =6.022×1023 atoms

The correct option is A $6.02 \times 10^{23}$
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: $\frac{4.25}{17} \times 4 \times 6.022 \times 10^{23} a t o m s$
= $6.022 \times 10^{23} a t o m s$