Chemistry, asked by urvashi616, 1 year ago

Crystalline CsBr has a bcc structure. Calculate the
unit cell edge length if the density of CsBr crystal is
4.24 g cm . (Atomic masses : Cs=133; Br = 80)​

Answers

Answered by IlaMends
14

Answer:

The  edge length of the unit cell of CsBr is 55.05 nm

Explanation:

CsBr has BCC structure.

Number of atom in unit cell of BCC (Z) = 42

Density of CsBr= 4.24 g/cm^3

Edge length = a

Atomic mass of CsBr(M) = 133 g/mol + 80 g/mol = 213 g/mol

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density

Z = number of atom in unit cell

M = atomic mass

(N_{A}) = Avogadro's number  

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'.

4.24 g/cm^3=\frac{2\times 213 g/mol}{6.022\times 10^{23} mol^{-1}\times a^{3}}

a = 5.505\times 10^{-8} m=55.05 nm

The  edge length of the unit cell of CsBr is 55.05 nm.

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