Elements A and B have fcc and bcc structures respectively with a unit cell edge length of $3 \mathrm{~A}^0$…
Elements A and B have fcc and bcc structures respectively with a unit cell edge length of $3 \mathrm{~A}^0$ for both elements. The no. of atoms in $210 \mathrm{gm}$ of $\mathrm{A}$ is equal to $594 \mathrm{gm}$ of $\mathrm{B}$. If density of $\mathrm{A}$ is $7 \mathrm{~g} \mathrm{~cm}^{-3}$, what is the density of $\mathrm{B}$ ?
$9.9 \mathrm{~g} \mathrm{~cm}^{-3}$
$4.5 \mathrm{~g} \mathrm{~cm}^{-3}$
$6.8 \mathrm{~g} \mathrm{~cm}^{-3}$
$11.2 \mathrm{~g} \mathrm{~cm}^{-3}$
Solution
$210 \mathrm{~g}$ of A contains $\left(\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{A}}} \times 210\right)$ no. of atoms $\mathrm{A}$ and $594 \mathrm{~g}$ of $\mathrm{B}$ contains $\left(\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{A}}} \times 594\right)$ no. of atoms $\mathrm{B}$.
$\left[\mathrm{M}_{\mathrm{A}}\right.$ and $\mathrm{M}_{\mathrm{b}}$ are the atomic weights of $\mathrm{A}$ and $\mathrm{B}$, respectively].
Now, $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{A}}} \times 210=\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{B}}} \times 594$
or $\frac{M_B}{M_A}=\frac{594}{210}$
Effective no. of $A$ atoms in $\mathrm{fcc}=4$ and that of $\mathrm{B}$ atoms in $\mathrm{bcc}=2$
$\mathrm{d}_{\mathrm{A}}=\frac{\mathrm{M}_{\mathrm{A}} \times 4}{\mathrm{~N}_{\mathrm{A}} \times \mathrm{a}^3}$ and $\mathrm{d}_{\mathrm{B}}=\frac{\mathrm{M}_{\mathrm{B}} \times 2}{\mathrm{~N}_{\mathrm{A}} \times \mathrm{a}^3} ; \mathrm{d}_{\mathrm{A}}=7 \mathrm{~g} / \mathrm{cm}^3$
$\therefore \frac{\mathrm{d}_{\mathrm{B}}}{\mathrm{d}_{\mathrm{A}}}=\frac{\mathrm{M}_{\mathrm{B}} \times 2}{\mathrm{M}_{\mathrm{A}} \times 4} ; \therefore \mathrm{d}_{\mathrm{B}}=7 \times \frac{594}{210} \times 2=9.9 \mathrm{~g} / \mathrm{cm}^3$