Question

AgCl has the same structure as that of NaCl. The edge length of a unit cell of AgCl is found to be $555$ pm and the density of AgCl is $5.561$ g cm${}^{-3}$. Find the percentage of sites that are unoccupied.

• A. $0.24\%$
• B. $2.4\%$
• C. $24\%$
• D. None

Answer 1

The correct option is A $0.24\%$

Chloride ions form fcc arrangement and silver ions occupy octahedral holes. The number of AgCl molecules (Z) in the unit cell is 4. The molar mass of AgCl is 143.32 g/mol

$d=\frac{Z\times M}{N_0\times a^3}$

$d=\frac{4\times 143.32}{6.023\times {10}^{23}\times (555\times {10}^{-10}{)}^3}$

$d=5.5677g/c{m}^3$

But the observed density is $5.561g/c{m}^3$

The percentage of sites that are unoccupied $=2\times \frac{5.5677-5.561}{5.5677}\times 100=0.24$%

Note: For each AgCl molecule that is absent, 2 sites are vacant. Hence, above expression is multiplied with 2.