Question

An element with $density6.8gc{m}^{-3}$, occurs in $bcc$ structure with cell edge length equal to $290pm$. Calculate the number of atoms present in $200g$ of the element.

• A. $2.4\times {10}^{42}$
• B. $1.2\times {10}^{42}$
• C. $1.2\times {10}^{24}$
• D. $2.4\times {10}^{24}$

Answer 1

Answer: (D)

$2.4\times {10}^{24}$

The expression for density (d) is given below:

$d=\frac{n\times M}{a^3\times N_A}$

For BCC,

$n=2$
$a=290$ pm $=290\times {10}^{-10}$ cm
$d=6.8$ g cm${}^{-3}$

$6.8=\frac{2\times M}{(290\times {10}^{-10}{)}^3\times 6.023\times {10}^{23}}$

$M=\frac{6.8\times 6.023\times {10}^{23}\times 24.4\times {10}^{-30}}{2}=50$

$\therefore$ Number of atoms present $=\frac{6.023\times {10}^{23}\times 200}{50}=2.4\times {10}^{24}$ atoms

Hence, the number of atoms present in $200$ g of the element is $2.4\times {10}^{24}$.