Answer
$\rho=8.50\frac{g}{cm^3}$
Work Step by Step
Given:
Atomic weight=141 g\mol
$a=b=0.35nm$ and $c=0.45 nm$
$\rho=\frac{nA}{{V_{c}}{N_{A}}}$
For BCC, n=2 .
$V_{c}=a\times b \times c$
$V_{c}=0.35\times 0.35 \times 0.45 \times 10^{-21} cm^3$
$N_{A}=6.023\times10^{23} \frac{atom}{mol}$
$\rho=\frac{2\times 141}{0.35\times 0.35 \times 0.45 \times 10^-21 \times6.023\times10^{23}}$
$\rho=8.50\frac{g}{cm^3}$
