Question

An element having a density of $8.24gc{m}^{-3}$ crystallizes as bcc lattice. The length of the side of the unit cell is $2.6\stackrel{o}{A}$. Calculate the number of atoms present in $468g$ of the element.

• A. $6.5\times {10}^{24}$
• B. $6.5\times {10}^{23}$
• C. $8.8\times {10}^{23}$
• D. $9.8\times {10}^{24}$

Answer 1

Answer: (A)

$6.5\times {10}^{24}$

As we know ,
$Density=\frac{z\times M}{N_A\times a^3}$
Where , given
$Density=8.24gc{m}^{-3}$
$N_A=6.022\times {10}^{23}$
$z=2$ (B.C.C lattice)
a = side length = $2.6A^0=2.6\times {10}^{-8}cm$
Putting the values of all this in above equation,
$8.24=\frac{2M}{6.022\times {10}^{23}\times {2.6}^3\times {10}^{-24}}$
After solving, we get
$M=43.6g$
Number of mole (n) = $\frac{givenmass}{molarmass}$
$n=\frac{468}{51.2}=10.73$
Number of atoms = n $\times N_A$
Number of atoms = $64.6\times {10}^{23}$
Hence , number of atoms present is 468g of the element = $6.46\times {10}^{24}$
Option A is correct.