Question

An element forms ccp lattice with a cell edgelength of 400 pm. The density of the elements
10 g cm 3. . the atomic mass of element will be
(1)65u
(2)54.2u
(3)96.3u
(4)205u​

Answer 1

Atomic mass of element is 96.3 u

Explanation:

The expression for the density of the unit cell is given below.  

$d=\frac{Z\times M}{a^3\times N_A}$

​where d = density  = $10g/cm^3$

Z = no of atoms = 4 (ccp)

M =  atomic mass = ?

a = edge length  = 400 pm = $400\times 10^{-10}cm$     $(1pm=10^{-10}cm)$

$N_A$ = Avogadros number = $6.023\times 10^{23}$

$10g/cm^3=\frac{4\times M}{(400\times 10^{-10})^3\times 6.023\times 10^{23}}$

$M=96.3u$

Atomic mass of element is 96.3 u

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Answer 2

Answer:

THE ANSWER IS 3) 96.3u

explanation:

it's very simple ....