⇒Solid State

Correct Option is (B)

$\begin{array}{ll}& \text{ Density }(\rho )=\frac{\mathrm{Mn}}{{\mathrm{a}}^3{\mathrm{N}}_{\mathrm{A}}}\\ \therefore & 21\mathrm{g}{\mathrm{cm}}^{-3}=\frac{\mathrm{M}\times 4}{36{\mathrm{cm}}^3{\mathrm{mol}}^{-1}}\\ \therefore & M=\frac{21\mathrm{g}{\mathrm{cm}}^{-3}\times 36{\mathrm{cm}}^3{\mathrm{mol}}^{-1}}{4}\\ & =189.00\mathrm{g}{\mathrm{mol}}^{-1}\end{array}$

Correct Option is (C)

$\mathrm{M}=100\mathrm{g}{\mathrm{mol}}^{-1},\mathrm{a}=400\mathrm{pm}=4\times {10}^{-8}\mathrm{cm}$

For BCC structure, $\mathrm{n}=2$ atoms

$\rho =\text{ ? }$

$\begin{array}{rl}\rho & =\frac{\mathrm{Mn}}{{\mathrm{a}}^3{\mathrm{N}}_{\mathrm{A}}}\\ & =\frac{100\mathrm{g}{\mathrm{mol}}^{-1}\times 2\text{ atoms }}{{(4\times {10}^{-8})}^3{\mathrm{cm}}^3\times 6.022\times {10}^{23}{\text{ atoms mol }}^{-1}}\\ & =\frac{200}{64\times {10}^{-24}\times 6.022\times {10}^{23}}\\ & =5.18\mathrm{g}{\mathrm{cm}}^{-3}\end{array}$

Correct Option is (C)

$\mathrm{M}=25\mathrm{g}{\mathrm{mol}}^{-1},\rho =3\mathrm{g}{\mathrm{cm}}^{-3}\text{, }$

For BCC structure, $\mathrm{n}=2$

Volume of unit cell $\left({\mathrm{a}}^3\right)=$ ?

$\begin{array}{rl}& \mathrm{M}=\frac{\rho \cdot {\mathrm{a}}^3{\mathrm{N}}_{\mathrm{A}}}{\mathrm{n}}\\ & \therefore {\mathrm{a}}^3=\frac{\mathrm{M}\times \mathrm{n}}{\rho \times {\mathrm{N}}_{\mathrm{A}}}\end{array}$

$\begin{array}{rl}\therefore a^3& =\frac{25\mathrm{g}{\mathrm{mol}}^{-1}\times 2\text{ atoms }}{3\mathrm{g}{\mathrm{cm}}^{-3}\times 6.022\times {10}^{23}\text{ atoms mol }}{}^{-1}\\ & =\frac{50\times {10}^{-23}}{18.066}=2.76\times {10}^{-23}{\mathrm{cm}}^3\end{array}$

Correct Option is (B)

$\rho =10\mathrm{g}{\mathrm{cm}}^{-3},\mathrm{a}=200\mathrm{pm}=2\times {10}^{-8}\mathrm{cm}$

For BCC structure, $\mathrm{n}=2$

$\begin{array}{rl}\mathrm{M}& =\frac{\rho {\mathrm{a}}^3{\mathrm{N}}_{\mathrm{A}}}{\mathrm{n}}\\ & =\frac{10\times {(2\times {10}^{-8})}^3\times 6.022\times {10}^{23}}{2}=\frac{10\times 8\times {10}^{-24}\times 6.022\times {10}^{23}}{2}\\ & =24.1\mathrm{g}{\mathrm{mol}}^{-1}\end{array}$

Correct Option is (C)

$\mathrm{M}=180\mathrm{g}\cdot {\mathrm{mol}}^{-1},\rho =18\mathrm{g}{\mathrm{cm}}^{-3}\text{, }$

For $\mathrm{BCC}$ crystal, $\mathrm{n}=2,\mathrm{a}=$ ?

$\begin{array}{rl}& \rho =\frac{\mathrm{M}\times \mathrm{n}}{{\mathrm{a}}^3\times {\mathrm{N}}_{\mathrm{A}}}\therefore {\mathrm{a}}^3=\frac{\mathrm{M}\times \mathrm{n}}{\rho \times {\mathrm{N}}_{\mathrm{A}}}\\ & \therefore {\mathrm{a}}^3=\frac{180\mathrm{g}{\mathrm{mol}}^{-1}\times 2\text{ atom }}{18\mathrm{g}{\mathrm{cm}}^{-3}\times 6.022\times {10}^{23}{\text{ atom mol}}^{-1}}\\ & \therefore {\mathrm{a}}^3=33.2\times {10}^{-24}{\mathrm{cm}}^3\\ & \therefore \mathrm{a}=\sqrt{33.2}\times {10}^{-8}\mathrm{cm}\end{array}$