Question

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?

Answer 1

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 =MassDensityMassDensity⇒208g7.2gcm−3⇒208g7.2gcm−3⇒28.89cm3⇒28.89cm3Number of unit cells =Total volumeVolume of a unit cellTotal volumeVolume of a unit cell⇒28.89cm32.389×10−23cm3⇒28.89cm32.389×10−23cm3⇒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

Answer 2

${\huge{\underline{\underline{\mathcal{\red{♡ANSWER♡}}}}}}$.

• ❤.. ${Volume of the unit cell}$=${(288pm)^3}$
• ❤..=${(288×10^-12 m)^-3 = (288×10^-10 cm)^3}$
• ❤..=${2.39×10^-23 cm^3}$
• ❤..${Volume of208 g of element}$
• ❤..$\frac{mass}{density}$=$\frac{208g}{7.2gcm^-3}$=${28.88cm^3}$
• ❤..${Number of unit cells in this volume}$

=$\frac{28.88cm^3}{2.39×10^-23Cm^3/unit cell}$=${12.08×10^23 unit cells}$

• ⭐.. 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.