An element has a bcc structure with unit cell edge length of 288 pm. How many unit cells and number of atoms are present in 200 g of the element? (1.16×10^24, 2.32×10^24)

An element has a bcc structure with unit cell edge length of 288 pm.

1.	An element has a bcc structure with unit cell edge length of 288 pm. How many unit cells and number of atoms are present in 200 g of the element? (1.16×10^24, 2.32×10^24)
Answer» tion:CHEMISTRYAn element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The DENSITY of the element is 7.2g/cm 3 . How many atoms present in 208G of the element. December 27, 2019avatarNavneet ArekarShareSaveANSWERVolume of UNIT cell =(288 pm) 3 =(288×10 −10 cm) 3 =2.389×10 −23 cm 3 Volume of 208 g of the element = DensityMass = 7.2208 =28.89 cm 3 Number of unit cells = Volume of a unit cellTotal Volume = 2.389×10 −23 28.89 =12.09×10 23 For a BCC structure, number of atoms per unit cell =2∴ Number of atoms present in 208 g = No. of atoms per unit cell × No. of unit cells =2×12.09×10 23 =24.18×10 23 =2.418×10 24