CsBr crystallizes in a body centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avogadro number being 6.02 × `10^(23)` mol–1, the density of CsBr is:-

Options

A. 8.25 `g//cm^3`
B. 4.25 `g//cm^3`
C. 42.5 `g//cm^3`
D. 0.425 `g//cm^3`

Answer

B

Explanation

Density of CsBr= `(Z xx M)/(a^3 xx N_0)`
Z=1(because 1 formula unit is presence)
M=molar mass of CsBr= 133 + 80 = 213
a=edge length of unit cell = 436.6 pm = `436.6xx 10^(-10)`cm
`:.` Density = `(1xx213)/((436.6xx10^(-10))^3xx6.02xx10^(23))=4.25 g//cm^3`