• # question_answer 68) The unit cell of a metallic element of atomic mass 108 and density $10.5\text{ }g/c{{m}^{3}}$ is a cube with edge length of 409 pm. The structure of the crystal lattice is- A) fcc                               B) bccC) hcp                              D) None of these

$\rho =\frac{Z\times M}{{{N}_{A}}\times {{a}^{3}}}$ Here, $M=108,\,\,{{N}_{A}}=6.023\times {{10}^{23}}$ $a=409\text{ }pm=4.09\,\,\times \,\,{{10}^{-}}^{8}cm,$ $p=10.5\text{ }g/c{{m}^{3}}$ Put on these values and solving we get. $Z\text{ }=\text{ }4\text{ }=\text{ }number\text{ }of\text{ }atoms\text{ }per\text{ }unit\text{ }cell$ So, the structure of the crystal lattice is fee.