$\begin{array}{1 1}7.30gcm^{-3}\\6.30gcm^{-3}\\4.30gcm^{-3}\\3.30gcm^{-3}\end{array} $

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Answer : $7.30gcm^{-3}$

In a body-centred cubic lattice ,atoms touch each other along the body-diagonal of the cube.Hence

$4r=\sqrt 3a$ or $r=\large\frac{\sqrt 3 a}{4}=\frac{(1.732)(287 pm)}{4}=$$124.27$pm.

The expression of density is $\rho=\large\frac{N}{a^3}\big(\frac{M}{N_A}\big)$

Substituting the values,we get

$\rho=\large\frac{2}{(287\times 10^{-10}cm)^3}\big(\frac{51.99gmol^{-1}}{6.023\times 10^{23}mol^{-1}}\big)$$=7.30gcm^{-3}$

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