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
B
bcc
C
hcp
D
None of these

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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