Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms.

Substance	Atomic Mass $(u)$	Density $(10^3\,kg\,m^{-3})$
Carbon (diamond)	$12.01$	$2.22$
Gold	$197.00$	$19.32$
Nitrogen (liquid)	$14.01$	$1.00$
Lithium	$6.94$	$0.53$
Fluorine (liquid)	$19.00$	$1.14$

To estimate the atomic radius $(r)$, we use the relation derived from the molar volume: $V_m = \frac{M}{\rho} = N_A \times \frac{4}{3} \pi r^3$, where $M$ is the molar mass, $\rho$ is the density, and $N_A = 6.023 \times 10^{23} \, mol^{-1}$ is Avogadro's number.
Thus, $r = \left( \frac{3M}{4 \pi \rho N_A} \right)^{1/3}$.
$\begin{array}{|l|c|} \hline \text{Substance} & \text{Radius } (\mathring{A}) \\ \hline \text{Carbon (diamond)} & 1.29 \\ \text{Gold} & 1.59 \\ \text{Nitrogen (liquid)} & 1.77 \\ \text{Lithium} & 1.73 \\ \text{Fluorine (liquid)} & 1.88 \\ \hline \end{array}$
$1$. For Carbon: $M = 12.01 \times 10^{-3} \, kg/mol$, $\rho = 2.22 \times 10^3 \, kg/m^3$. $r = [3 \times 12.01 \times 10^{-3} / (4 \pi \times 2.22 \times 10^3 \times 6.023 \times 10^{23})]^{1/3} \approx 1.29 \, \mathring{A}$.
$2$. For Gold: $M = 197.00 \times 10^{-3} \, kg/mol$, $\rho = 19.32 \times 10^3 \, kg/m^3$. $r = [3 \times 197.00 \times 10^{-3} / (4 \pi \times 19.32 \times 10^3 \times 6.023 \times 10^{23})]^{1/3} \approx 1.59 \, \mathring{A}$.
$3$. For Nitrogen (liquid): $M = 14.01 \times 10^{-3} \, kg/mol$, $\rho = 1.00 \times 10^3 \, kg/m^3$. $r = [3 \times 14.01 \times 10^{-3} / (4 \pi \times 1.00 \times 10^3 \times 6.023 \times 10^{23})]^{1/3} \approx 1.77 \, \mathring{A}$.
$4$. For Lithium: $M = 6.94 \times 10^{-3} \, kg/mol$, $\rho = 0.53 \times 10^3 \, kg/m^3$. $r = [3 \times 6.94 \times 10^{-3} / (4 \pi \times 0.53 \times 10^3 \times 6.023 \times 10^{23})]^{1/3} \approx 1.73 \, \mathring{A}$.
$5$. For Fluorine (liquid): $M = 19.00 \times 10^{-3} \, kg/mol$, $\rho = 1.14 \times 10^3 \, kg/m^3$. $r = [3 \times 19.00 \times 10^{-3} / (4 \pi \times 1.14 \times 10^3 \times 6.023 \times 10^{23})]^{1/3} \approx 1.88 \, \mathring{A}$.