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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

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