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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