Question

An element having the density of $6.8$ g cm${}^{-3}$ occurs in bcc structure with cell edge length of $290$ pm. The number of atoms present in $200$ g of the element is:

• A. $24\times {10}^{23}$
• B. $24\times {10}^{24}$
• C. $24\times {10}^{22}$
• D. $24\times {10}^{20}$

Answer 1

Answer: (A)

$24\times {10}^{23}$

As we know,
$\text{Density}=\frac{n\times \text{Atomic mass}}{N_o\times a^3}$ or $\frac{n\times A}{6.023\times {10}^{23}\times (290\times {10}^{-10{)}^3}}=6.8$
$\therefore$ $A=50$
$\therefore$ Number of atoms in $200$ g $=\frac{6.023\times {10}^{23}\times 200}{50}=24\times {10}^{23}$