Chemistry, asked by vaibhavd1701, 1 year ago

An element has a BCC structure with a cell edge of 288pm.The density of the element is 7.2g/cm3.How many atoms are present in 208g of the element?

Answers

Answered by arc2003
2
Volume of unit cell =(288pm)3(288pm)3⇒(288×10−10cm)3⇒(288×10−10cm)3⇒2.389×100−23cm3⇒2.389×100−23cm3Volume of 208g of the element =MassDensityMassDensity208g7.2gcm−3208g7.2gcm−3⇒28.89cm3⇒28.89cm3Number of unit cells =Total volumeVolume of a unit cellTotal volumeVolume of a unit cell28.89cm32.389×10−23cm328.89cm32.389×10−23cm3⇒12.09×1023⇒12.09×1023For a bcc structure,number of atom per unit cell = 2Number of atoms present in 208g=No. of atoms per unit cell ×× No. of unit cells⇒2×12.09×1023⇒2×12.09×1023⇒24.18×1023⇒24.18×1023⇒2.418×1024
Answered by SugaryGenius
2

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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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