An element has a body-centred cubic (bcc) structure with a cell edge of 288 pm. The density of the element is 7.2 g/cm3 .How many atoms are present in 104 g of the element ?
Answers
Answer:
An element has a body Centred cubic structure with a cell edge of 288 pm
An element has a bcc structure with a celledge of `288` pm. The density of the element is `7.2 g cm^(-3)`. How many atins are present in `208 g` of the element? Doubtnut provides Chemistry Solutions for Class 9, class 10, class 11, class 12 and IIT-JEE aspiran
Explanation:
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
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