Question

An element has a body centred cubic structure with cell edge of 288 pm. The density of the element is 7.2 g/cm3. How many atoms are present in 208 g of the element.​

Answer 1

Answer :

• No. of atoms present in 208 g of the element = 24.17 × 10²³ atoms.

Step-by-step explanation :

Given that,

• Edge length of unit cell, a = 288 pm = $\bold{288\times{}10^{-10}\:cm}$

• Density of the element, $\bold{\rho=7.2\:g/cm^3}$

$\hrulefill$

For BBC structure, Z = 2

Now, we know, $\bold{\rho=\dfrac{Z\times{}M}{a^3\times{}N_0}}$

Substituting the values in the expression, we get,

$\bold{7.2\:g\:cm^{-3}=\dfrac{2\times{}M}{(288\times{}10^{-10}\:cm)^3\times{}(6.02\times{}10^{23}\:mol^{-1})}}$

or $\bold{M=51.8\:g\:mol^{-1}}$

By mole concept, 51.8 g of the element contains = 6.02 × 10²³ atoms.

∴ 208 g of the element contains,

$\implies\bold{\dfrac{6.02\times{}10^{23}}{51.8}\times{}208\:atoms}$

$\implies\bold{24.17\times{}10^{23}\:atoms\:(Ans.)}$

Answer 2

Given :

An element has BCC structure

Edge, a=288pm

Density of element , d =7.2 g/cm³

To find ;

How many atoms are present in 208 g of the element.

Theory :

•BCC (body centred cubic structure)

1. 8 corner atoms contribute one atom per unit cell.
2. Centre atom contributes one atom per unit cell.
3. So, Total 1+1=2 atoms per unit cell

$\bf\:Z_{effective}=8\times\frac{1}{8}$

•Density of unit cell

$\sf\:Density=\dfrac{mass\:of\:unit\: cell}{Volume\:of\:unit\:cell}$

$\implies\:\sf\:Density=\dfrac{ZM}{a{}^{3}N_{A}}$

where , $\sf\:N_{A}=Avaogadro\:Number$

and M=molar mass

and $\sf\:Z=Z_{effective}$

Solution:

Volume of unit cell =$a{}^{3}$

$=\sf\:(288\times\:10{}^{-12}){}^{3}$

$=2.39\times10{}^{-23}\:cm{}^{3}$

We know that for BCC structure

$\sf\:Z_{effective}=2$

$\sf\:Density=\dfrac{ZM}{a{}^{3}N_{a}}$

Now put the values

$\implies \sf 7.2 = \frac{2 \times M}{(288 \times 10 {}^{ - 12}) {}^{3} \times 6.023 \times 10 {}^{23} }$

$\implies \sf 7.2 = \frac{ \cancel{2} \times M}{2.39 \times 10 {}^{ - 23} \times \cancel{6} \times 10 {}^{23} }$

$\implies \sf 7.2 = \frac{M\times \cancel{10 {}^{23}} }{2.39 \times 3 \times \cancel{10 {}^{23} }}$

$\implies \sf M = 7.2 \times 2.39 \times 3 \: g$

$\implies \sf M =51.62g$

By mole concept,51.62 g element contains =$N_{A}atoms$

$\implies \sf288g \: element \: contains = \frac{6.023 \times 10 {}^{23} \times 288 }{51.6}$

$\implies\:\sf\:288g\:element\:contains=2.41\times10{}^{24}\: atoms$