An element occurs in bcc structure with cell edge 300pm. The density of the element is 5.2g/cm3 .How many atoms of the element does 200g of the element contain?
Answers
cubic unit cell lattice parameter a = 288 pm = 288 x10-8 cm
Volume V = a3 =2.39X10-23cm3
Density d = 7.2g/cm3
NA = Avogadro constant = 6.022x10²³
Molecular mass M =?
We know that
Density d = ZM/NA X a3
M = dxNA x a3/Z
On Substituting values
M= 7.2g/cm3 x(6.022x10²³)X (6.022x10²³)/2
= 51.8gmol-1
51.8 g of element contains 6.022X1023
208g of this element contains=?
= 6.022X1023X208/51.8
=2.42X1024 atoms.
Explanation:
Volume of unit cell =(288 pm) ³
=(288×10 ⁻¹° cm) ³
=2.389×10⁻²³cm ³
Volume of 208 g of the element = Density x Mass = 7.2 x 208
=28.89 cm ³
Number of unit cells = Volume of a unit cell /Total 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
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