`Li` forms a body-centred cubic lattice. If the edge of the cube is `3.5 xx 10^(-10)m` and the density is `5.3 xx 10^(2) kg m^(-3)`, calculate the percentage occupancy of `Li` metal.

Options

A. `25.78%`
B. `1.54g//cm^(2)`
C. `3.4g//c^(3)`
D. `5.4g//cm^(3)`

Answer

D

Explanation

We have theoretical density `=(zM)/(NV)=(zM)/(N(a^(3)))`
For a b.c.c. lattice`,z=2` and given that `a=3.50xx10^(-10)` and `M=7xx10^(-3)kg//mol e`
`d_(cell)=(2xx(7xx10^(-3)))/(6.022xx10^(23)xx(3.50xx10^(-10))^(3))=5.42xx10^(2)kgm^(-3)`