Which has maximum number of atoms?

Solution :
$${\text{TIPS/Formulae:}}$$
$$\eqalign{ & {\text{Atomic weight in gms}}\,{\text{ = }}\,{\text{6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}\,{\text{atoms}}\,{\text{ = 1}}\,{\text{Mole}}\,{\text{atoms}} \cr & \left( {\text{i}} \right){\text{Number}}\,{\text{of}}\,{\text{atoms}}\,{\text{in}}\,{\text{24}}g\,{\text{of}}\,C \cr & {\text{ = }}\frac{{24}}{{12}} \times 6.023 \times {10^{23}} = 2 \times 6.023 \times {10^{23}}\,{\text{atoms}} \cr & {\text{ = 2}}\,{\text{mole}}\,{\text{atoms}} \cr & \left( {{\text{ii}}} \right)\,{\text{Number}}\,{\text{of}}\,{\text{atoms}}\,{\text{in}}\,{\text{56}}g\,{\text{of}}\,Fe \cr & {\text{ = }}\frac{{56}}{{56}} \times 6.023 \times {10^{23}} = 6.023 \times {10^{23}}{\text{atom}} \cr & = 1\,{\text{mole}}\,{\text{atoms}} \cr & \left( {{\text{iii}}} \right)\,{\text{Number}}\,{\text{of}}\,{\text{atoms}}\,{\text{in}}\,{\text{27g}}\,{\text{of}}\,Al \cr & {\text{ = }}\frac{{27}}{{27}} \times 6.023 \times {10^{23}} = 6.023 \times {10^{23}}{\text{atom}} \cr & {\text{ = 1}}\,{\text{mole}}\,{\text{atoms}} \cr & \left( {{\text{iv}}} \right)\,{\text{Number}}\,{\text{of}}\,{\text{atoms}}\,{\text{in}}\,{\text{108}}g\,{\text{of}}\,Ag \cr & {\text{ = }}\frac{{108}}{{108}} \times 6.023 \times {10^{23}} = 6.023 \times {10^{23}}\,{\text{atom}} \cr & = 1\,{\text{mole}}\,{\text{atoms}} \cr & \therefore \,24g\,{\text{of}}\,C\,{\text{has}}\,{\text{maximum}}\,{\text{number}}\,{\text{of}}\,{\text{atoms}}. \cr} $$