Question

Silver forms ccp lattice and X-ray studies of its crystals show that the edge length of its unit cell is 408.6 pm. Calculate the density of silver (Atomic mass = 107.9 u).
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Answer 1

Solution :

Silver forms cubic close packing (ccp) , therefore the number of silver atoms per unit cell z=4

Molar mass of silver M=107.9 gmol−1 = 107.9×10−3kgmol−1

Edge length of unit cell a=408.6 pm = 408.6×10−12

Density d=(z.M)/(a3.NA)

d=4×(107.9×10−3kgmol−1)/(408.6×10−12m)3(6.022×1023mol−1)

=10.5×103kgm−3

d=10.5gcm−3

Answer 2

Answer :

• Density of silver = $\bold{10.5\:g\:cm^{-3}}$

Step-by-step explanation :

Given that,

• Atomic mass(M) = $\bold{107.9\:g\:mol^{-1}}$

• Edge length(a) of its unit cell = $\bold{408.6\:pm=408.6\times{}10^{-10}\:cm}$

$\hrulefill$

Also, for ccp lattice (which is equivalent to fcc lattice), Z = 4 atoms/unit cell.

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

Applying the given values, we get,

$\implies\bold{\rho=\dfrac{4\:atoms\times{}107.9\:g\:mol^{-1}}{(408.6\times{}10^{-10}\:cm)^3\times{}(6.022\times{}10^{23}\:atoms\:mol^{-1})}}$

$\implies\bold{\rho=10.5\:g\:cm^{-3}}$

∴ Density of silver = $\bold{10.5\:g\:cm^{-3}}$