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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 $6.02 \times 10^{23}$ mol$^{-1}$, the density of CsBr is :-
A
42.5 g/cm$^3$
B
0.425 g/cm$^3$
C
8.25 g/cm$^3$
D
4.25 g/cm$^3$
Detailed Solution
Density of a crystal: $d = \dfrac{n \times M}{V \times N_A}$, where $n$ is the number of formula units per unit cell.
CsBr has the CsCl type structure: $Br^-$ ions at the corners and $Cs^+$ at the body centre, so the number of formula units per unit cell is $n = 1$.
Molar mass: $M = 133 + 80 = 213$ g mol$^{-1}$
Edge length: $a = 436.6$ pm $= 436.6 \times 10^{-10}$ cm
Volume: $V = a^3 = (436.6 \times 10^{-10})^3 = 83.22 \times 10^{-24}$ cm$^3$
$d = \dfrac{1 \times 213}{83.22 \times 10^{-24} \times 6.02 \times 10^{23}}$
$d = \dfrac{213}{50.1} = 4.25$ g/cm$^3$
CsBr has the CsCl type structure: $Br^-$ ions at the corners and $Cs^+$ at the body centre, so the number of formula units per unit cell is $n = 1$.
Molar mass: $M = 133 + 80 = 213$ g mol$^{-1}$
Edge length: $a = 436.6$ pm $= 436.6 \times 10^{-10}$ cm
Volume: $V = a^3 = (436.6 \times 10^{-10})^3 = 83.22 \times 10^{-24}$ cm$^3$
$d = \dfrac{1 \times 213}{83.22 \times 10^{-24} \times 6.02 \times 10^{23}}$
$d = \dfrac{213}{50.1} = 4.25$ g/cm$^3$