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Question: > Total number of atoms present in 34 g of \(NH_{3}\)is > >...

Total number of atoms present in 34 g of NH3NH_{3}is

A

4×10234 \times 10^{23}

B

4.8×10214.8 \times 10^{21}

C

2×10232 \times 10^{23}

D

48×102348 \times 10^{23}

Answer

48×102348 \times 10^{23}

Explanation

Solution

(4) : No. of moles in 34 g of lNH3=3417=2NH_{3} = \frac{34}{17} = 2

No. of moleculsc=2×6.023×1023= 2 \times 6.023 \times 10^{23}

No. of atoms in one molecule of NH3NH_{3}=4

No. of atoms in 2 molecules of NH3NH_{3}

=4×2×6.023×1023=48.18×1023= 4 \times 2 \times 6.023 \times 10^{23} = 48.18 \times 10^{23}