Which has more number of atoms,100,g of sodiu | vedclass

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(D) The number of atoms in a given mass of an element is calculated using the formula: $	ext{Number of atoms} = \frac{	ext{Given mass}}{	ext{Atomic

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$100\,g$ of sodium

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(D) The number of atoms in a given mass of an element is calculated using the formula: $	ext{Number of atoms} = \frac{	ext{Given mass}}{	ext{Atomic mass}} 	imes N_A$,where $N_A$ is Avogadro's number $(6.022 	imes 10^{23} 	ext{ mol}^{-1})$.For $100\,g$ of Sodium $(Na)$:$	ext{Number of atoms} = \frac{100}{23} 	imes 6.022 	imes 10^{23} \approx 2.618 	imes 10^{24}$ atoms.For $100\,g$ of Iron $(Fe)$:$	ext{Number of atoms} = \frac{100}{56} 	imes 6.022 	imes 10^{23} \approx 1.075 	imes 10^{24}$ atoms.Comparing the two values,$2.618 	imes 10^{24} > 1.075 	imes 10^{24}$.Therefore,$100\,g$ of sodium contains a greater number of atoms than $100\,g$ of iron.

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

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