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$\mathrm{KBr}$ has rock salt type structural arrangements and has a density of $3.70 \mathrm{~g} / \mathrm{cm}^3$. The edge length of the unit cell is approximately [molecular weight of $\mathrm{KBr}=120 \mathrm{~g} / \mathrm{mol}$ ]
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The correct answer is:
$6 \times 10^{-8} \mathrm{~cm}$
$\mathrm{KBr}$ has fcc type structural arrangement.
$\therefore \quad$ Effective no. of atoms (z) in a unit cell $=4$
$$
\begin{array}{ll}
\therefore & \mathrm{d}=\frac{\mathrm{ZM}}{\mathrm{N}_{\mathrm{A}} \mathrm{a}^3} \text { where } \mathrm{a}=\text { the edge length } \\
\therefore & \mathrm{a}^3=\frac{\mathrm{ZM}}{\mathrm{N}_{\mathrm{A}} \mathrm{d}}=\frac{4 \times 120}{6.023 \times 10^{23} \times 3.7} \mathrm{~cm}^3 \\
\therefore & \mathrm{a}=6 \times 10^{-8} \mathrm{~cm}
\end{array}
$$
$\therefore \quad$ Effective no. of atoms (z) in a unit cell $=4$
$$
\begin{array}{ll}
\therefore & \mathrm{d}=\frac{\mathrm{ZM}}{\mathrm{N}_{\mathrm{A}} \mathrm{a}^3} \text { where } \mathrm{a}=\text { the edge length } \\
\therefore & \mathrm{a}^3=\frac{\mathrm{ZM}}{\mathrm{N}_{\mathrm{A}} \mathrm{d}}=\frac{4 \times 120}{6.023 \times 10^{23} \times 3.7} \mathrm{~cm}^3 \\
\therefore & \mathrm{a}=6 \times 10^{-8} \mathrm{~cm}
\end{array}
$$
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