How much does the mass of NH3 contain same number of atom as are present in 4 gram atom of Oxygen?
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Answer:
Let moles of NH3 = x
No. of atoms in 1 molecule of NH3 = (1+3) = 4
Therefore,
No. of atoms in x moles NH3 = x * N * 4 …..eq(i)
(where N is Avogadro no.)
No of moles in 4g O2 = 4/32 (Mol. Wt of O2 = 32)
No of atoms in 1 molecule of O2 = 2
Therefore,
No. of atoms in 4g O2 = (4/32) * N * 2 …..eq(ii)
Equating eq (i) & (ii),
x * N * 4 = (4/32) * N * 2
Therefore, x = 16
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Therefore for =16
I hope this answer is right
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