Find the maximum number of atoms in the follo | vedclass

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(C) The number of atoms is calculated as: $	ext{Number of atoms} = \frac{	ext{mass}}{	ext{molar mass}} 	imes N_A 	imes 	ext{atomicity}$.$(i)$ $1

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$(iii)$

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(C) The number of atoms is calculated as: $	ext{Number of atoms} = \frac{	ext{mass}}{	ext{molar mass}} 	imes N_A 	imes 	ext{atomicity}$.$(i)$ $1 \ g$ $Au$: $\frac{1}{197} 	imes N_A \approx 0.00507 	imes N_A$ atoms.$(ii)$ $1 \ g$ $Na$: $\frac{1}{23} 	imes N_A \approx 0.04347 	imes N_A$ atoms.$(iii)$ $1 \ g$ $Li$: $\frac{1}{7} 	imes N_A \approx 0.14285 	imes N_A$ atoms.$(iv)$ $1 \ g$ $Cl_2$: $\frac{1}{71} 	imes N_A 	imes 2 \approx 0.02816 	imes N_A$ atoms.Comparing the values,$1 \ g$ $Li$ has the maximum number of atoms.

Find the maximum number of atoms in the following:
$(i)$ $1 \ g$ $Au_{(s)}$
$(ii)$ $1 \ g$ $Na_{(s)}$
$(iii)$ $1 \ g$ $Li_{(s)}$
$(iv)$ $1 \ g$ $Cl_{2(g)}$

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Find the maximum number of atoms in the following:$(i)$ $1 \ g$ $Au_{(s)}$$(ii)$ $1 \ g$ $Na_{(s)}$$(iii)$ $1 \ g$ $Li_{(s)}$$(iv)$ $1 \ g$ $Cl_{2(g)}$