Question

21. An element A crystallises in fcc structure. 200 g of
this element has 4.12 x 1024 atoms. The density
of A is 7.2 g cm Calculate the edge length of the
unit cell.
(1) 26.97 x 10-24 cm (2) 299.9 pm
(3) 5 x 10-12 cm
(4) 2.99 cm​

Answer 1

The edge length of the  unit cell is, (2) 299.9 pm

Explanation :

Formula used :  

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$ .............(1)

where,

$\rho$ = density  = $7.2g/cm^3$

Z = number of atom in unit cell (for FCC = 4)

M = atomic mass  = 200 g

$(N_{A})$ = Avogadro's number  or number of atoms = $4.12\times 10^{24}$

a = edge length of unit cell

Now put all the values in above formula (1), we get

$7.2g/cm^3=\frac{4\times (200g/mol)}{(4.12\times 10^{24}mol^{-1}) \times (a^3)}$

$a=2.999\times 10^{-8}cm=299.9\times 10^{-10}cm=299.9pm$

Thus, the edge length of the  unit cell is, 299.9 pm

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