Question

The density of iridium is 22.4 g/cm^3. The unit cell of iridium is fcc. Calculate the radius of iridium atom. Molar mass of iridium is 192.2 g/mol​​

Answer 1

Answer:

1.35 * 1/10⁸ cm

...................

.

Answer 2

Answer:

The radius of iridium atom is 136 pm.

Step-by-step-explanation:

We have given that,

Density of iridium = 22.4 g/cm³

Crystal structure of iridium = fcc

Molar mass of iridium = 192.2 g/mol

We have to find the radius of iridium atom.

Now,

fcc crystal structure means face centred cubic unit cell.

In this structure, there are 8 atoms in 8 corners and 6 atoms at 6 face centres.

Only 1 / 8 th part of atoms at corner contribute to 1 unit cell.

No. of atoms at corner = 1 / 8 * 8 = 1

And, only half part of atoms present at face centres contribute to 1 unit cell.

No. of atoms at face centre = 1 / 2 * 6 = 3

∴ Total no. of atoms in 1 unit cell = 1 + 3 = 4

Now, we know that,

$\displaystyle{\boxed{\pink{\sf\:Mass\:in\:grams\:=\:\dfrac{Molar\:mass}{Avogadro's\:number}\:}}}$

$\displaystyle{\implies\sf\:Mass\:of\:iridium\:atom\:=\:\dfrac{192.2\:g/mol}{6.022\:\times\:10^{23}\:mol^{-1}}}$

$\displaystyle{\implies\sf\:Mass\:of\:iridium\:atom\:=\:\dfrac{\cancel{192.2}}{\cancel{6.022}\:\times\:10^{23}}\:g}$

$\displaystyle{\implies\sf\:Mass\:of\:iridium\:atom\:=\:\dfrac{\cancel{96.1}}{\cancel{3.011}\:\times\:10^{23}}\:g}$

$\displaystyle{\implies\sf\:Mass\:of\:iridium\:atom\:=\:\dfrac{31.916}{10^{23}}\:g}$

$\displaystyle{\implies\sf\:Mass\:of\:iridium\:atom\:=\:31.916\:\times\:10^{-\:23}}$

$\displaystyle{\implies\sf\:Mass\:of\:iridium\:atom\:=\:31.92\:\times\:10^{-\:23}\:g}$

As there are 4 atoms of iridium,

Mass of 4 atoms = 4 * 31.92 * 10⁻²³

⇒ Mass of 4 atoms = 127.68 * 10⁻²³ g

∴ Mass of unit cell = 127.68 * 10⁻²³ g

Now,

$\displaystyle{\boxed{\blue{\sf\:Density\:of\:unit\:cell\:=\:\dfrac{Masa\:of\:unit\:cell}{Volume\:of\:unit\:cell}\:}}}$

The unit cell is cubic in shape. Let the edge length of unit cell be "a" cm.

$\displaystyle{\therefore\:\sf\:22.4\:=\:\dfrac{127.68\:\times\:10^{-23}}{a^3}}$

$\displaystyle{\implies\:\sf\:a^3\:=\:\dfrac{\cancel{127.68}\:\times\:10^{-23}}{\cancel{22.4}}}$

$\displaystyle{\implies\:\sf\:a^3\:=\:5.7\:\times\:10^{-23}}$

$\displaystyle{\implies\:\sf\:a^3\:=\:5.7\:\times\:10\:\times\:\dfrac{10^{-\:23}}{10}}$

$\displaystyle{\implies\:\sf\:a^3\:=\:57\:\times\:10^{-\:24}}$

$\displaystyle{\implies\:\sf\:a\:=\:\sqrt[\sf\:3]{57\:\times\:10^{-\:24}}}$

$\displaystyle{\implies\:\boxed{\purple{\:\sf\:a\:=\:3.848\:\times\:10^{-\:8}\:cm}}}$

For fcc structure,

$\displaystyle{\boxed{\green{\sf\:Radius\:of\:atom\:=\:\dfrac{a}{2\:\sqrt{2}}\:}}}$

$\displaystyle{\implies\:\sf\:Radius\:of\:atom\:=\:\dfrac{\cancel{3.848}\:\times\:10^{-\:8}}{\cancel{2}\:\sqrt{2}}}$

$\displaystyle{\implies\:\sf\:Radius\:of\:atom\:=\:\dfrac{1.924\:\times\:10^{-\:8}}{\sqrt{2}}}$

$\displaystyle{\implies\:\sf\:Radius\:of\:atom\:=\:\dfrac{\cancel{1.924}\:\times\:10^{-\:8}}{\cancel{1.414}}}$

$\displaystyle{\implies\:\sf\:Radius\:of\:atom\:=\:1.36\:\times\:10^{-\:8}\:cm}$

$\displaystyle{\implies\:\sf\:Radius\:of\:atom\:=\:1.36\:\times\:100\:\times\:\dfrac{10^{-\:8}}{10^2}}$

$\displaystyle{\implies\:\sf\:Radius\:of\:atom\:=\:136\:\times\:10^{-\:10}}$

$\displaystyle{\implies\underline{\boxed{\red{\:\sf\:Radius\:of\:atom\:=\:136\:pm\:}}}}$

∴ The radius of iridium atom is 136 pm.