Answer

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**Hint:**For a one mole of a substance of atomic radius r and density $\rho $ having mass M is the mass is equal to volume of the atom times density. All the other parameters are given and the radius of the atom means the size would be found out.

**Complete step by step answer:**

We know that for a one mole of a substance with atomic radius r and density $\rho $ having mass M . Let us suppose the atoms to be spherical in nature.

Avogadro’s number $ = 6.022 \times {10^{23}}$

We know that density is equal to mass divided by volume. Then $\rho = \dfrac{m}{V}$

$ \Rightarrow M = \rho \times V$

Where M is the mass, $\rho $ is the density and $V$ is the volume of the spherical atom

Then, $M = {N_A}\dfrac{4}{3}\pi {r^3}\rho $

$ \Rightarrow r = {(\dfrac{{3M}}{{4{N_A}\pi \rho }})^{\dfrac{1}{3}}}$

Now for carbon atom mass M = $12.01 \times {10^{ - 3}}\;kg$

Density, $\rho = 2.22 \times {10^3}\;kg\;{m^{ - 3}}$

$\therefore r = {(\dfrac{{3 \times 12.01 \times {{10}^{ - 3}}}}{{4\pi \times 2.22 \times {{10}^3} \times 6.023 \times {{10}^{23}}}})^{\dfrac{1}{3}}}$

On doing the above calculations , we get

$r = 1.29\;{A^ \circ }$

So the radius of the carbon atom is $1.29\;{A^ \circ }$

Now for gold M $ = 197.00 \times {10^{ - 3}}\;kg$

$\rho = 19.32 \times {10^3}\;kg\;{m^{ - 3}}$

$\therefore r = {(\dfrac{{3 \times 197 \times {{10}^{ - 3}}}}{{4\pi \times 19.32 \times {{10}^3} \times 6.023 \times {{10}^{23}}}})^{\dfrac{1}{3}}}$

$ \Rightarrow r = 1.59\;{A^ \circ }$

Hence the radius of the gold atom is $1.59\;{A^ \circ }$

Now for liquid nitrogen M$ = 14.01 \times {10^{ - 3}}\;kg$

density, $\rho = 1.00 \times {10^3}\;kg\;{m^{ - 3}}$

$ \Rightarrow r = 1.77\;{A^ \circ }$

Hence the radius of the liquid nitrogen atom is $1.77\;{A^ \circ }$

$\therefore r = {(\dfrac{{3 \times 14.01 \times {{10}^{ - 3}}}}{{4\pi \times 1.00 \times {{10}^3} \times 6.023 \times {{10}^{23}}}})^{\dfrac{1}{3}}}$

Now for lithium atom M$ = 6.94 \times {10^{ - 3}}\;kg$

Density, $\rho = 0.53 \times {10^3}\;kg\;{m^{ - 3}}$

$\therefore r = {(\dfrac{{3 \times 6.94 \times {{10}^{ - 3}}}}{{4\pi \times 0.53 \times {{10}^3} \times 6.023 \times {{10}^{23}}}})^{\dfrac{1}{3}}}$

$ \Rightarrow r = 1.73\;{A^ \circ }$

Hence the radius of lithium atom is $1.73\;{A^ \circ }$

Now for liquid fluorine atom M $ = 19.00 \times {10^{ - 3}}\;kg$

Density, $\rho = 1.14 \times {10^3}\;kg\;{m^{ - 3}}$

$\therefore r = {(\dfrac{{3 \times 19 \times {{10}^{ - 3}}}}{{4\pi \times 1.14 \times {{10}^3} \times 6.023 \times {{10}^{23}}}})^{\dfrac{1}{3}}}$

$ \Rightarrow r = 1.88\;{A^ \circ }$

Hence the radius of the liquid fluorine atom is $1.88\;{A^ \circ }$.

**Note:**For the above question the proper conversions should be made correctly . Also for a unit cell with z number of atoms, M mass the density is calculated according to the formula

$\rho = \dfrac{{ZM}}{{{N_A}{a^3}}}$

Where Z is the number of atoms, M is the molecular mass , ${N_A}$ is the Avogadro number and a is the edge length of the unit cell.

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