Question

An element having atomic mass 60.2 has a face centred cubic unit cell. The edge length of the unit cell is 100 pm. Find out density of the element. (Na = 6.02x10^23)

Answer 1

Answer:

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Explanation:

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Answer 2

The density of the element in the 6.23 $\frac{g}{cm^{3}}$.

Explanation:

The having atomic mass $60.2$ has a face center cubic unit cell

$F_{cc}=(Ze_{ff}=4)$

$\frac{Ze_{ff}*A_{w}}{a^{3}*NA}$

$=\frac{4*60}{(100)^{3}*(6.023*10^{23}*10^{-10}}$

$=6.23\frac{g}{cm^{3}}$.