Question

An element having atomic mass 60.2 has a face
centred cubic unit cells .The edgelength of the
unit cell is 100 pm. Find out density of the element
(NA = 6.02 x 10^23)

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Answer 1

Explanation:

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Answer 2

Answer:

The density of the element is 400 g/cm³.

Explanation:

The density of the element will be calculated by the following formula,

$\mathrm{d=\frac{Z\times M}{N_a\times a^3} }$                 (1)

Where,

d=density of the element

Z= total number of atoms in a unit cell

M= atomic weight of the element

$\mathrm{N_a}$=avogadro constant=6.02×10²³

a=edge length of the unit cell

From the question we have,

The atomic weight of the element(M)=60.2

The edge length of the unit cell(a)=100 pm=100×10⁻¹⁰cm

Equation (1) yields; when all relevant values are entered into it,

$\mathrm{d=\frac{4\times 60.2}{6.02\times 10^{23}\times (100\times 10^{-10})^3 }}$

$\mathrm{d=400 \ g /cm^3 }}$

So, the density of the element is 400 g/cm³.

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