Question

CsBr crystallises in a body centred cubic lattice. the unit cell length is 436.6 pm . given that the atomic mass of Cs = 133 and that of Br= 80 amu, what will be the density of CsBr?

Answer 1

$$\begin{gathered} CsBrisabcclattice.Sonoofparticlesinbcc\left(Z\right)=2 \\ Density, \\ \rho =\frac{Z\times M}{a^3\times N_A} \\ Where \\ Z=noofparticles=2 \\ M=molecularmass=133+80=213g \\ a=436.6pm=436.6\times {10}^{-10}cm \\ {N}_A=avogadronumber=6.022\times {10}^{23} \\ Density, \\ \rho =\frac{Z\times M}{a^3\times N_A}=\frac{2\times 213}{{\left(436.6\times {10}^{-10}\right)}^3\times 6.022\times {10}^{23}} \\ =8.499g/c{m}^3 \end{gathered}$$