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  • An element crystallizes in a b.c.c lattice with cell edge of 500pm. The density of the element is 7.5g cm-3. How many atoms are present in 300g of the element?     &nb

An element crystallizes in a b.c.c lattice with cell edge of 500pm. The density of the element is 7.5g cm-3. How many atoms are present in 300g of the element?

 

 

 

 
 
 
 
 

Answers (1)

Mass = 300g

Density = 7.5g cm-3

Edge = 500pm

Volume of unit cell = (500)^{3}

                              = 1.25\times 10^{-22}cm^{3}

Volume of 300g = \frac{mass}{density} = \frac{300}{7.5}=40cm^{3}

No. of unit cell in 40cm3 = \frac{40}{1.25\times 10^{-22}}= 32\times 10^{22} unit cells

Total number of atoms in 300g = 2\times 32\times 10^{22}= 64\times 10^{22}atoms

Posted by

Sumit Saini

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