The number of molecules in $4.25 g$ of ammonia is approximately?

3.5 \times 10^{23}

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0.5 \times 10^{23}

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2.5 \times 10^{23}

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1.5 \times 10^{23}

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Solution

The correct option is A $1.5 \times 10^{23}$
No of moles = $\frac{W e i g h t}{g M w t}$
= $\frac{4.25}{17}$
= $0.25$

No of Moles $\times N_{A}$ = Number of Molecules

Number of molecules = $0.25 \times 6.023 \times 10^{23}$

= $1.5 \times 10^{23}$

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Similar questions

The number of molecules in 4.25 g of ammonia is approximately

THe number of molecules in 4.25 g of ammonia is:
(a) $1.0 \times 10^{23}$ (b) $1.5 \times 10^{23}$
(c) $2.0 \times 10^{23}$ (d) $3.5 \times 10^{23}$

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}

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