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(C) $(i).$ $1 \,g$ of $Au_{(s)} = \frac{1}{197} \,mol$ of $Au_{(s)} = \frac{6.022 	imes 10^{23}}{197} \approx 3.06 	imes 10^{21}$ atoms of $Au_{(s)}

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$(iii)$

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(C) $(i).$ $1 \,g$ of $Au_{(s)} = \frac{1}{197} \,mol$ of $Au_{(s)} = \frac{6.022 	imes 10^{23}}{197} \approx 3.06 	imes 10^{21}$ atoms of $Au_{(s)}$.$(ii).$ $1 \,g$ of $Na_{(s)} = \frac{1}{23} \,mol$ of $Na_{(s)} = \frac{6.022 	imes 10^{23}}{23} \approx 26.2 	imes 10^{21}$ atoms of $Na_{(s)}$.$(iii).$ $1 \,g$ of $Li_{(s)} = \frac{1}{7} \,mol$ of $Li_{(s)} = \frac{6.022 	imes 10^{23}}{7} \approx 86.0 	imes 10^{21}$ atoms of $Li_{(s)}$.$(iv).$ $1 \,g$ of $Cl_{2(g)} = \frac{1}{71} \,mol$ of $Cl_{2(g)}$. Since $1 \,mol$ of $Cl_2$ contains $2 \,mol$ of $Cl$ atoms,number of atoms $= \frac{2 	imes 6.022 	imes 10^{23}}{71} \approx 16.96 	imes 10^{21}$ atoms of $Cl$.Comparing the values,$1 \,g$ of $Li_{(s)}$ has the largest number of atoms.

Which one of the following will have the largest number of atoms?
$(i)$ $1 \,g$ $Au_{(s)}$
$(ii)$ $1 \,g$ $Na_{(s)}$
$(iii)$ $1 \,g$ $Li_{(s)}$
$(iv)$ $1 \,g$ of $Cl_{2(g)}$

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Find the maximum number of atoms in the following:
$(i)$ $1 \ g$ $Au_{(s)}$
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The largest number of molecules is in

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Which one of the following will have the largest number of atoms?$(i)$ $1 \,g$ $Au_{(s)}$$(ii)$ $1 \,g$ $Na_{(s)}$$(iii)$ $1 \,g$ $Li_{(s)}$$(iv)$ $1 \,g$ of $Cl_{2(g)}$