An element crystallising in body centred cubic lattice has an edge length of 500 pm. If its density is `4 g cm^(-3)`, the atomic mass of the element (in g `mol^(-1)`) is (consider `N_(A) = 6 xx 10^(23)`)

Options

A. 100
B. 250
C. 125
D. 150

Answer

D

Explanation

Relation used,
Density `(rho) = (Z xx M)/(a^(3) xx N_(A))`
where, Z = contribution factor
M = molar mass of particle
a = edge length
`rho` = density of crystal
Given, `N_(A)` = Avogadro.s Number
`= 6 xx 10^(23)`
`rho = 4g//cm^(3)`
= 2 (crystal has bcc structure)
`therefore M = (rho xx a^(3) xx N_(A))/(Z)`
`= (4 xx (500 xx 10^(-10))^(3) xx 6 xx10^(23))/(2)`
`M = (4 xx (5)^(3) xx 6 xx 10^(-1))/(2)`
`= (4 xx 125 xx 6 xx 10^(-1))/(2)`
`M = (300)/(2) = 150 g mol^(-1)`
`therefore` The atomic mass of the element (in g `mol^(-1)`) is 150.