Question

An elemental crystal has a density of $8570$ kg m${}^{-3}$. The packing efficiency is $0.68$. If the closest distance between neighbouring atoms is $2.86$ $\mathring{A}$, the mass of one atom is:

$(1$ amu $=1.66\times {10}^{-27}$ kg$)$

• A. $186.0$ amu
• B. $94.6$ amu
• C. $46.5$ amu
• D. $43.2$ amu

Answer 1

The correct option is B $94.6$ amu

Since the packing efficiency is $0.68$, the given element crystallizes in BCC lattice. For BCC lattice, $z=2.$

In BCC closest distace $=2r$ and $4r=a\sqrt{3}.$

Here, $r=$ radius of ion/atom and $a=$ edge length.

$⟹a=3.30\mathring{A}=3.3\times {10}^{-10}m$

$\text{Density, d}=\frac{z\times \frac{M}{1000}}{N_A\times a^3}$

$8750=\frac{2\times \frac{M}{1000}}{6.02\times {10}^{23}\times (3.30\times {10}^{-10}{)}^3}$

$M=\frac{8750\times 6.02\times {10}^{23}\times 35.937\times {10}^{-30}\times 1000}{2}$

$M=94.6g/mol$
$\therefore$ Mass of 1 atom $=94.6amu$