Q.

Number of atoms present in 4.25g of NH3 is:

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a

6.023×1023

b

4×6.023×1023

c

1.7×1024

d

4.25×6.023×1023

answer is A.

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Detailed Solution

No. of moles of ammonia = 4.25/17 = 0.25 molNo. of molecules of ammonia =  6.023×1023×0.25One molecule contains a total 4 atoms.Thus, total no. of atoms in ammonia =6.023×1023×0.25×4=6.023×1023
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Number of atoms present in 4.25g of NH3 is: