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Q.

The number of atoms in 4.25 g of NH3 is approximately

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a

1 × 1023

b

2 × 1023

c

4 × 1023

d

6 × 1023

answer is D.

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Detailed Solution

∵ 17 g NH3 contains 6 × 1023 molecules of NH3∴ 4.25 g NH3 contains = 6 × 102317 × 4.25∴ No. of atoms = 6 × 1023 × 4.2517 × 4=6 × 1023

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