Question

An element having the fcc lattice structure has a cell edge of 120 pm. The density of the element is $6.8g/c{m}^3$. How many atoms are present in 408 g of the element?

• A. ^{$1.73\times {10}^{26}$}
• B. $1.22\times {10}^{26}$
• C. $1.38\times {10}^{26}$
• D. $1.73\times {10}^{25}$

Answer 1

The correct option is B $1.22\times {10}^{26}$

Volume of unit cell = $(120pm{)}^3=(120\times {10}^{-12}{)}^3{m}^3=({12}^3\times {10}^{-33})m^3$
Volume of 408g of the element $=\frac{mass}{density}=\frac{408}{6.8}=6\times {10}^{-5}{m}^3$
So, number of unit cells presents in 408 g of the elements $=\frac{6\times {10}^{-5}}{{12}^3\times {10}^{-33}}=3.472\times {10}^{25}$ unit cells.
Since each fcc unit cell consist of $\frac{7}{2}$ atoms,
Therefore the total number of atoms presents in 408 g of the given element
$=\frac{7}{2}\times 3.472\times {10}^{25}=1.22\times {10}^{26}$ atoms