Silver crystallises in FCC structure. If density of silver is $10.51 g . {c m}^{- 3}$ . Calculate the volume of unit cell. [Atomic mass of silver $(A g) = 108 g m^{- 1}$ ]

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Solution

Mass of $1$ atom of $A g = \frac{A t o m i c m a s s o f A g}{N_{A}}$
$= \frac{108}{6.033 \times 10^{23}}$
$= 17.03 \times 10^{- 23}$
Mass of unit cell $=$ Atoms in unit cell $\times$ Mass of $1$ atom
$= 4 \times 17.03 \times 10^{- 23}$
$= 71.71 \times 10^{- 23} g$
Volume of unit cell $= \frac{M a s s o f u n i t c e l l}{D e n s i t y}$
$= \frac{71.71 \times 10^{- 23}}{10.51}$
$= 68.24 \times 10^{- 24}$

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