Question

Given that the density of KBr is $2.48$ g cm${}^{-3}$ and it crystallizes in FCC, the edge length of its unit cell (in pm) is :

Answer 1

We apply density formula $\rho =\frac{ZM}{N_0{a}^3}$ and solve this problem easily.
Now by putting values, we get $2.48=\frac{3\times 58}{(6.022\times {10}^{23})\times (a{)}^3}⟹a=541$ pm