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An element makes fcc structure which edge length is 200 pm. If 24×10^23 atoms presence in 200g of element then find the density by using D=(Z×M) /(a^3 ×na)?
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An element makes fcc structure which edge length is 200 pm. If 24×10^2...
For Fcc

Z=4

N=24×1023 atoms

a=200 pm

=2×10−8

d=a3×N2×M​=8×10−24×24×10234×200​

=41.67 g/cm3

The density of 200 g of this element is =41.67 g/cm3
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An element makes fcc structure which edge length is 200 pm. If 24×10^23 atoms presence in 200g of element then find the density by using D=(Z×M) /(a^3 ×na)?
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An element makes fcc structure which edge length is 200 pm. If 24×10^23 atoms presence in 200g of element then find the density by using D=(Z×M) /(a^3 ×na)? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about An element makes fcc structure which edge length is 200 pm. If 24×10^23 atoms presence in 200g of element then find the density by using D=(Z×M) /(a^3 ×na)? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An element makes fcc structure which edge length is 200 pm. If 24×10^23 atoms presence in 200g of element then find the density by using D=(Z×M) /(a^3 ×na)?.
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