AgX crystallizes in rock salt like structure in which $X^{-}$ occupy FCC lattice site. The density of AgX crystal found to be $\frac{40}{6.022} g / c m^{3}$ and the edge length of the unit cell is 600 pm. What will be the mass (in grams) of 1 mole unit cell if all the atoms along one body diagonal of each unit cell have been removed. [Given: Atomic weight of Ag = 108 g/mol, $N_{A} = 6.022 \times 10^{23}$ ]

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answer is 729.

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Detailed Solution

$M_{A g X} = \frac{ρ \times N_{A} \times a^{3}}{4} = \frac{\frac{40}{6.022} \times 6.022 \times 10^{23} \times {(6 \times 10^{- 8})}^{3}}{4} = 216 g / m o l e$

$∴ M_{X} = 216 - 108 = 108 g / m o l e$

If all atoms along one body diagonal have been removed, the formula of unit cell
$= A g_{4 - 1} X_{4 - \frac{1}{4}} = A g_{3} X_{\frac{15}{4}}$
$∴$ Mass of 1 mole of these unit cell $= 108 \times 3 + 108 \times \frac{15}{4} = 729.00$