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Question

The unit cell of a metallic element of atomic mass 108 and density 10.5 $${ g/cm }^{ 3 }$$ is a cube with edge length of 409 pm. The structure of the crystal lattice  is:


A
fcc
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B
bcc
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C
hcp
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D
None of these
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Solution

The correct option is A fcc

We know that,

Density-

$$d=\dfrac{Z\times M}{a^3×NA}$$


Given :

Molar Mass$$ (M) = 108 g/ mol$$

Density $$(d) = 10.5 g/cm^3$$

Edge Length$$(a) = 409 pm$$


$$Z=\dfrac{d\times a^3×N_A}{M} $$

Value of $$N_A $$ is $$N_A=6.023\times 10^{23}$$


$$Z=\dfrac{10.5\times (409\times 10^{−10}cm^3)^3\times 6.023\times 10^{23}}{ 108}=4$$


Number of atoms $$= 4$$


Hence the element is packed in FCC structure.


Chemistry

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