Question

An element occurs in the bcc structure with a cell edge of $288$ pm. The density of this element is $7.2$ g/cc. How many atoms are present in $208$ g of this element?

• A. $24.16\times {10}^{25}$
• B. $24.16\times {10}^{23}$
• C. $24.16\times {10}^{24}$
• D. $24.16\times {10}^{26}$

Answer 1

Answer: (B)

$24.16\times {10}^{23}$

The formula for density is given below:

$d=\frac{ZM}{N_A{a}^3}$, where $z$ is the number of atoms present in a lattice type.

Substituing the values, we get

$\frac{2\times M}{{6\times 10}^{23}\times {(288\times {10}^{-10})}^3}=7.2$ g/cc $⟹M=52$ g/mol

Number of moles of element $=\frac{208}{52}=4$

Number of atoms $=N_A\times 4$ $=24.16\times {10}^{23}$

Option B is correct.