Chemistry, asked by V2139, 10 months ago

an element has a bcc sptructure with a cell edge of 288 pm how many unit cells and number of atoms are present in 200 gm of the element?

Answers

Answered by majnu14377
1

Explanation:

200gm

200pm

200×6.023×10^23/200

6.023×10^23

Answered by SugaryGenius
6

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  • ❤.. {Volume of the unit cell}={(288pm)^3}
  • ❤..={(288×10^-12 m)^-3 = (288×10^-10 cm)^3}
  • ❤..={2.39×10^-23 cm^3}
  • ❤..{Volume of208 g of element}
  • ❤..\frac{mass}{density}=\frac{208g}{7.2gcm^-3}={28.88cm^3}
  • ❤..{Number of unit cells in this volume}
  • =\frac{28.88cm^3}{2.39×10^-23Cm^3/unit cell}={12.08×10^23 unit cells}
  • ⭐.. Since each{bcc}cubic unit cell contains {2} atoms,therefore,the total number of atoms in {208g = 2}{(atoms/unit cell)×12.08×10^23} unit cells={24.16×10^23} atoms.
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