Solution:
Number of atoms \(=\) Number of moles
\(\times\) Avogadro's number \(\left(N_{A}\right)\) Number of atoms in \(24 \mathrm{~g} \mathrm{C}=\frac{24}{12} \times N_{A}=2 N_{A}\)
Number of atoms in \(56 \mathrm{~g}\) of \(\mathrm{Fe}=\frac{56}{56} N_{A}=N_{A}\)
Number of atoms in \(27 \mathrm{~g}\) of \(\mathrm{Al}=\frac{27}{27} N_{A}=N_{A}\)
Number of atoms in \(108 \mathrm{~g}\) of \(\mathrm{Ag}=\frac{108}{108} N_{A}=N_{A}\)
Hence, \(24 \mathrm{~g}\) of carbon has the maximum number of atoms.
