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Question: Ag forms CCP lattices with edge length \(408.6pm\). Calculate density of silver (Atomic mass \(=107....

Ag forms CCP lattices with edge length 408.6pm408.6pm. Calculate density of silver (Atomic mass =107.9μ=107.9\mu)

Explanation

Solution

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%74\% of available space is occupied. The structure of ccp lattice is as follows:

Complete step by step answer:
First we will discuss about the statement given in question,
The edge length of the unit cell =408.6pm=408.6×1012m=408.6×1010cm= 408.6pm =408.6\times {{10}^{\\_12}} m = 408.6\times {{10}^{\\_10}}cm.
Now we will calculate, the density of silver =mass of unit cellvolume of unit cell= \dfrac{\text{mass of unit cell}}{\text{volume of unit cell}}
The volume of unit cell =a3= {{a}^{3}}
It mean (408.6×1024)3cm{{\left( 408.6\times {{10}^{\\_24}} \right)}^{3}}cm = 68.27×1024cm368.27\times {{10}^{\\_24}}c{{m}^{3}}
To calculate the mass of unit cell, the formula is
=atomic mass of silverAvogadro’s number= \dfrac{\text{atomic mass of silver}}{\text{Avogadro’s number}}
Atomic mass =107.9μ= 107.9 \mu (given)
Avogadro’s number =6.023×1023per mol= 6.023\times {{10}^{23}} \text{per mol}
Mass of unit cell =107.9μ6.023×1023per mol= \dfrac{107.9\mu}{ 6.023\times {{10}^{23}} \text{per mol}}
(6.023×1023per mol6.023\times {{10}^{23}} \text{per mol} is the value of Avogadro's number)
=17.91×1023per mol= 17.91\times {{10}^{\\_23}} \text{per mol}
CCP unit cell has four atoms per unit cell,
Mass of unit cell =4×17.91×1023=71.64×1023g= 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 =71.64×1023g= 71.64\times {{10}^{\\_23}}g (calculated)
Volume of unit cell =68.27×1024cm3= 68.27\times {{10}^{\\_24}}c{{m}^{3}}
Density of silver =mass of unit cell volume of unit cell= \dfrac{\text{mass of unit cell}}{\text{ volume of unit cell}}
=71.64×1023g68.27×1024cm3=1.042×101=10.42g/cm3= \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 =10.42g/cm3= 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%74\% space in the crystal is filled. The coordination number is 12 for the cubic close packing.