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Ag forms CCP lattices with edge length $408.6pm$. Calculate density of silver (Atomic mass $=107.9\mu$)

Answer
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Hint: CCP LATTICES stands for cubic close packing and packing covered octahedral or‘d’ voids. The arrangement of packing is ABCABCABC……… type and is known as face-centered cubic. In this type of packing, $74\%$ of available space is occupied. The structure of ccp lattice is as follows:
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Complete step by step answer:
First we will discuss about the statement given in question,
The edge length of the unit cell $= 408.6pm =408.6\times {{10}^{\_12}} m = 408.6\times {{10}^{\_10}}cm$.
Now we will calculate, the density of silver $= \dfrac{\text{mass of unit cell}}{\text{volume of unit cell}}$
The volume of unit cell $= {{a}^{3}}$
It mean ${{\left( 408.6\times {{10}^{\_24}} \right)}^{3}}cm$ = $68.27\times {{10}^{\_24}}c{{m}^{3}}$
To calculate the mass of unit cell, the formula is
$= \dfrac{\text{atomic mass of silver}}{\text{Avogadro’s number}}$
Atomic mass $= 107.9 \mu$ (given)
Avogadro’s number $= 6.023\times {{10}^{23}} \text{per mol}$
Mass of unit cell $= \dfrac{107.9\mu}{ 6.023\times {{10}^{23}} \text{per mol}}$
($6.023\times {{10}^{23}} \text{per mol}$ is the value of Avogadro's number)
$= 17.91\times {{10}^{\_23}} \text{per mol}$
CCP unit cell has four atoms per unit cell,
Mass of unit cell $= 4\times 17.91\times {{10}^{\_23}} = 71.64\times {{10}^{\_23}}g$
Now we will calculate the density of silver,
Mass of unit cell $= 71.64\times {{10}^{\_23}}g$ (calculated)
Volume of unit cell $= 68.27\times {{10}^{\_24}}c{{m}^{3}}$
Density of silver $= \dfrac{\text{mass of unit cell}}{\text{ volume of unit cell}}$
$= \dfrac{71.64\times {{10}^{\_23}}g}{68.27\times {{10}^{\_24}}c{{m}^{3}}} = 1.042\times {{10}^{1}} = 10.42 g/c{{m}^{3}}$
So, the density of silver $= 10.42 g/c{{m}^{3}}$

Note:
Iron, nickel, copper, gold, silver and aluminum crystallize in ccp structure. These types of close packing are highly efficient and $74\%$ space in the crystal is filled. The coordination number is 12 for the cubic close packing.