Answer
$\rho=10.22\frac{g}{cm^3}$; the theoretical value and the experimental value are the same.
Work Step by Step
given ${r}=0.1363 nm$ Atomic weight=95.94g\mol ,and experimental density ${\rho_{exp}}=10.2\frac{g}{ cm^3}$
$\rho=\frac{nA}{{V_{c}}{N_{A}}}$
for BCC, n=2 and $V_{c}=(\frac{4r}{\sqrt{3}})^3$
$N_{A}=6.023\times10^{23} \frac{atom}{mol}$ is Avagadro's number.
$\rho=\frac{2\times95.94}{(\frac{4r}{\sqrt{3}})^3\times6.023\times10^{23}}$
$\rho=10.22\frac{g}{cm^3}$
