Question

Which has more number of atoms: (a) 100 g of sodium or (b) 100 g of iron?

(Given: atomic mass of Na= 23 u, Fe= 56 u)

Answer 1

Step-1: Calculation of the number of atoms in $100\mathrm{g}$ of Sodium $\left(\mathrm{Na}\right)$

According to the mole concept, $1\mathrm{mole}=6.022\times {10}^{23}\mathrm{atoms}\mathrm{of}\mathrm{the}\mathrm{substance}=\mathrm{Gram}\mathrm{atomic}\mathrm{weight}\mathrm{of}\mathrm{that}\mathrm{substance}$

• Therefore, $23\mathrm{g}$ of sodium contains= $6.022\times {10}^{23}$ atoms of sodium
• $1\mathrm{g}$ of sodium contains= $\frac{6.022\times {10}^{23}}{23}$ atoms of sodium
• $100\mathrm{g}$ of sodium contains= $\frac{6.022\times {10}^{23}}{23}\times 100=26.18\times {10}^{23}$ atoms of sodium

Step-2: Calculation of the number of atoms in $100\mathrm{g}$ of Iron $\left(\mathrm{Fe}\right)$

According to the mole concept, $1\mathrm{mole}=6.022\times {10}^{23}\mathrm{atoms}\mathrm{of}\mathrm{the}\mathrm{substance}=\mathrm{Gram}\mathrm{atomic}\mathrm{weight}\mathrm{of}\mathrm{that}\mathrm{substance}$

• Therefore,$56\mathrm{g}$ of iron contains= $6.022\times {10}^{23}$ atoms of iron
• $1\mathrm{g}$ of iron contains= $\frac{6.022\times {10}^{23}}{56}$ atoms of iron
• $100\mathrm{g}$ of iron contains= $\frac{6.022\times {10}^{23}}{56}\times 100=10.75\times {10}^{23}$ atoms of iron

Step-3: Compare the number of atoms in $100\mathrm{g}$ in sodium and iron obtained in step-1 and 2.

It is found that $100\mathrm{g}$ of sodium has $26.18\times {10}^{23}$atoms of sodium whereas $100\mathrm{g}$ of iron has $10.75\times {10}^{23}$atoms of iron. Therefore, $100\mathrm{g}$ of sodium has more atoms than in $100\mathrm{g}$ of iron.