Which one of the following will have largest number of atoms

Solution:

(i). $1 \mathrm{~g}$ of $\mathrm{Au}(\mathrm{s})=\frac{1}{197} \mathrm{~mol}$ of $\mathrm{Au}(\mathrm{s})$

$=\frac{6.022 \times 10^{23}}{197}$ atoms of $A u(s)$

$=3.06 \times 10^{21}$ atoms of $A u(s)$

(ii). $1 \mathrm{~g}$ of $\mathrm{Na}(\mathrm{s})=\frac{1}{23}$ $\mathrm{mol}$ of $\mathrm{Na}(\mathrm{s})$

$=\frac{6.022 \times 10^{23}}{23}$ atoms of $\mathrm{Na}(\mathrm{s})$

$=0.262 \times 10^{23}$ atoms of $N a(s)$

$=26.2 \times 10^{21}$ atoms of $\mathrm{Na}(\mathrm{s})$

(iii). $1 \mathrm{~g}$ of $\mathrm{Li}(\mathrm{s})=\frac{1}{7} \mathrm{~mol}$ of $\mathrm{Li}(\mathrm{s})$

$=\frac{6.022 \times 10^{23}}{7}$ atoms of $\mathrm{Li}(\mathrm{s})$

$=0.86 \times 10^{23}$ atoms of $\mathrm{Li}(\mathrm{s})$

$=86.0 \times 10^{21}$ atoms of $\mathrm{Li}(\mathrm{s})$

(iv). $1 \mathrm{~g}$ of $\mathrm{Cl}_{2}$ (g) $=\frac{1}{71}$ mol of $\mathrm{Cl}_{2}$ (g)

(Molar mass of $\mathrm{Cl}_{2}$ molecule $=35.5 \times 2=71 \mathrm{~g} \mathrm{~mol}^{-1}$ )

$=\frac{6.022 \times 10^{23}}{71}$ atoms of $\mathrm{Cl}_{2}(\mathrm{~g})$

$=0.0848 \times 10^{23}$ atoms of $\mathrm{Cl}_{2}(g)$

$=8.48 \times 10^{21}$ atoms of $\mathrm{Cl}_{2}(\mathrm{~g})$

Hence, 1 g of Li (s) will have the largest number of atoms.