Chemistry, asked by somamajumder888, 11 months ago

an element occurs in BCC structure with a edge length of 288 pm. the density of the element is 7.2 gm cm-3. how many atoms of the element does 208 g of the element contain? ​

Answers

Answered by SugaryGenius
3

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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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