Li' forms a body centered cubic lattice. If the lattice constant is `3.5xx10^(-10)m` and the experimental density is `5.30xx10^(3) kgm^(-3)`, calculate the percentage occupancy of Li metal.

Explanation

We have,
theoretical density `=(zm)/(NV) = (zM)/(Na^(3_))`
For a b.c.c. lattice: z = 2 and given that `a = 3.50 xx 10^(-10)` m and `M = 7 xx 10^(-3)` kg/mole.
`d_("cal") = 2 xx (7 xx 10^(-3))/(6.022 xx 10^(23) xx (3.50 xx 10^(-10))^(3))`
`=5.42 xx 10^(2) kg m^(-3)`
`therefore` percentage occupancy `=(rho_("exp")/rho_("cal")) xx 100`
`=(5.30 xx 10^(2))/(5.42 xx 10^(2)) xx 100 = 97.78%`