Answer
396.9k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)