Question -Assume a hypothetical cubic crystal lattice, named JEE-centered cubic (jcc) with the following characteristics: I. An atom is present at all the corners of the cube II. An atom is present at the center of two pairs of opposite faces III. An atom is present at the center of all the edges of the cube IV. One atom is present at its body-center An element having the jcc lattice structure has a cell edge of 120 pm. The density of the element is 6.8 g/cm3. How many atoms are present in 408 g of the element?
Options:
A .  1.73×1023
B .  2.43×1023
C .  2.43×1026
D .  1.73×1026
Answer: Option C : C Volume of unit cell =(120pm)3=(120×10−12)3m3=(123×10−33)m3 Volume of 408 g of the element =massdensity=4086.8=60cm3=6×10−5m3 So, number of unit cells present in 408 g of the elements =6×10−5123×10−33=3.472×1025 unit cells Since each jcc unit cell consist of 7 atoms, therefore the total number of atoms presents in 408 g of the given element =7×3.472×1025=2.43×1026atoms
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