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.

Options

A. 0.8778
B. 0.9987
C. 0.9778
D. 0.9412

Answer

C

Explanation

We have ,
theoretical density ` = (zM)/(NV) = (zM)/(N(a^3))`
for a bcc lattice z = 2 and given that a ` = 3.50 xx 10^(-10) m`
`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)`
percentage occupancy ` = (rho_(exp) )/(rho_(cal)) xx 100`
` = (5.30 x 10^2)/(5.42 xx 10^2) xx 100 = 97.78%`