Which among the following elements has the hi | vedclass

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(B) The number of atoms is calculated by the formula: $	ext{Number of atoms} = \frac{	ext{Given mass}}{	ext{Molar mass}} 	imes N_A$. Since the mas

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$Na_{(s)}$

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(B) The number of atoms is calculated by the formula: $	ext{Number of atoms} = \frac{	ext{Given mass}}{	ext{Molar mass}} 	imes N_A$. Since the mass is $1 \ g$ for all,the number of atoms is inversely proportional to the molar mass. $I$. $1 \ g \ Fe = \frac{1}{56} 	imes N_A \approx 0.0178 	imes N_A$ atoms. $II$. $1 \ g \ Au = \frac{1}{197} 	imes N_A \approx 0.0051 	imes N_A$ atoms. $III$. $1 \ g \ Na = \frac{1}{23} 	imes N_A \approx 0.0435 	imes N_A$ atoms. $IV$. $1 \ g \ Cu = \frac{1}{63.5} 	imes N_A \approx 0.0157 	imes N_A$ atoms. Comparing the values,$Na$ has the smallest molar mass,therefore it has the highest number of atoms.

Which among the following elements has the highest number of atoms in $1 \ g$ each? (Atomic masses: $Au=197, Na=23, Cu=63.5, Fe=56$)

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