Question

Which has more number of atoms, $100$ grams of sodium or $100$grams of iron (given, atomic mass of $\mathrm{Na}=23\mathrm{u}$, $\mathrm{Fe}=56\mathrm{u}$$100\mathrm{g}$)?

Answer 1

Step 1: Given data

Amount of sodium = $100\mathrm{g}$

Amount of iron = $100\mathrm{g}$

Atomic mass of sodium = $23\mathrm{u}$

Atomic mass of iron = $56\mathrm{u}$

Step 2: Formula used

One mole of any substance(molecular weight in grams) contains $6.022\times {10}^{23}$ number of atoms. This is known as the Avogadro's number.

Therefore, unitary method is used to calculate the number of atoms in $100\mathrm{g}$ of sodium and iron.

Step 3: Calculation of the number of atoms in 100g sodium

Weight of one mole of sodium = $23\mathrm{g}$

Using Avagadro's number,

Number of atoms $23\mathrm{g}$ of sodium = $6.022\times {10}^{23}$

Number of atoms $1\mathrm{g}$ of sodium = $\frac{6.022\times {10}^{23}}{23}$

Number of atoms $100\mathrm{g}$ of sodium = $\frac{6.022\times {10}^{23}}{23}\times 100=2.6182\times {10}^{24}$

Step 4: Calculation of the number of atoms in 100g iron

Weight of one mole of iron = $56\mathrm{g}$

Using Avagadro's number,

Number of atoms $56\mathrm{g}$ of iron = $6.022\times {10}^{23}$

Number of atoms $1\mathrm{g}$ of iron = $\frac{6.022\times {10}^{23}}{56}$

Number of atoms $100\mathrm{g}$ of iron = $\frac{6.022\times {10}^{23}}{56}\times 100=1.0753\times {10}^{24}$

Hence, $100$ grams of sodium has more atoms than $100$grams of iron.