An element has BCC structure with unit cell edge length of 288 picometre how many unit cells and number of atoms are present in 200 grams of the element?
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CHEMISTRY
An 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, 2019avatar
Navneet Arekar
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ANSWER
Volume of unit cell =(288 pm)
3
=(288×10
−10
cm)
3
=2.389×10
−23
cm
3
Volume of 208 g of the element =
Density
Mass
=
7.2
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
Answer:
no. of atoms in 200 g element is 1.16 X 10²⁴
no. of unit cells in 200 g element is 5.80 X 10²³
