Silver crystallises in a face-centred cubic (FCC) lattice with edge length a = 408 pm — JEE Chemistry
Silver crystallises in a face-centred cubic (FCC) lattice with edge length $a = 408$ pm. Calculate the radius of the Ag atom and the density of silver. (Atomic mass = 108 g/mol)
1 Answer
VAVidaara Admin
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· 26d ago
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For FCC: atoms touch along face diagonal:
$$4r = \sqrt{2}\,a \Rightarrow r = \frac{\sqrt{2} \times 408}{4} = \frac{577}{4} = 144 pm$$
FCC has $Z = 4$ atoms per unit cell:
$$\rho = \frac{Z \times M}{N_A \times a^3} = \frac{4 \times 108}{6.022\times10^{23} \times (408\times10^{-10})^3}$$
$$= \frac{432}{6.022\times10^{23} \times 6.79\times10^{-23}} = \frac{432}{40.88} = 10.57 g/cm^3$$
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Discussion (3)
AS
This finally made it click for me — thank you!
SG
I solved it a slightly different way and got the same answer, good sign.
AB
Brilliant explanation, the substitution step is what I kept missing.