An element with molar mass `2.7 xx 10^(2) kg mol^(-1)` forms a `2.7 xx 10^(3) kg^(-3)`, what is the nature of the cubic unit cell?

Options

A. sc
B. fcc
C. bcc
D. end centred

Answer

B

Explanation

To solve out such problems, first calculate the number of atoms present in the unit cell
(Z) `=(d xxa^(3)xxN_(A))/(M)` and then find the type of the
cubic unit cell as Z= 1 for scc, 2 for bcc and 4 for fcc as .
Density (d) `=(ZxxM)/(a^(3)xxN_(A)` so, Z `=(d xxa^(3)xxN_(A))/(M)`
Given M `=2.7xx10^(-2)` kg `mol^(-1)`
`a=405` pm `405xx10^(-12)m=4.05xx10^(-10)m`
d `=2.7xx10^(3)` kg `m^(-3)`
`N_(A)=6.022xx10^(23)mol^(-1)`
Hence,
`Z=((2.7xx10^(23))(4.05xx10^(-10)m)^(3)xx(6.022xx10^(23)mol^(-1)))/((2.7xx10^(-2)kg" mol"^(-1)))`
Z`=3.99=4`
Since, there are four atoms per unit cell the cubic unit cell must be face-centred