Question

$AgCl$ has the same structure as that of $NaCl$. The edge length of unit cell of $AgCl$ is found to be $555$ pm and the density of $AgCl$ is $5.561gc{m}^{-3}$. If the percentage of sites that are unoccupied is $x$, the value of $100x$ is:

Answer 1

The molar mass of AgCl is 143.32 g/mol.

$d=\frac{ZM}{N{a}^3}$

Here, d is the density of AgCl, Z is the number of formula units present in one unit cell, M is the molar mass of AgCl, a is the edge length of AgCl unit cell and N is the Avogadro's number.

$5.561gc{m}^{-3}=\frac{Z\times 143.32}{6.023\times {10}^{23}\times (555\times {10}^{-10}{)}^3}$

$Z=3.995194$

Thus, the percentage of the sites that are unoccupied is $x=\frac{4-3.995194}{4}\times 100=0.12$ %

$100x=100\times 0.12=12$