Question

an element having atomic mass 60.2 has a face centred cubic unit cells.The edge length of the unit cell is 100pm .Find out the density of the element.(Na=6.02*10+23)​

Answer 1

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Answer 2

Answer:

The density of the given element of molar mass $60.2u$ calculated is $4\times 10^{8} g/m^{3}$.

Explanation:

Given data,

The atomic mass/molecular mass of the given element, M = $60.2u$

The unit cell is a face-centred cubic unit cell ( fcc ).

The unit cell's given edge length, a = $100pm$

Convert pm in m

• $1pm = 10^{-12}m$
• $100pm = 100\times 10^{-12}m$ = $10^{-10}m$

The value of the Avogadro's number = $6.02\times 10^{23}$

The density of the given element, d =?

From the formula of the density for a unit cell given below, we can find out the density of the element:

• d = $\frac{Z\times M}{a^{3}\times N_{A} }$

Here, Z = number of atoms per unit cell

And for fcc, Z = 4

Therefore,

• d = $\frac{4\times 60.2}{(10^{-10})^{3}\times 6.02\times 10^{23} }$
• d = $\frac{240.8}{10^{-30}\times 6.02\times 10^{23} }$
• d = $\frac{240.8\times 10^{7} }{6.02 }$
• d = $40\times 10^{7}$ or $4\times 10^{8}g/m^{3}$

Hence, the density of the given element having molar mass $60.2u$ calculated is $4\times 10^{8} g/m^{3}$.