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If an element having atomic number 96 crystallises in cubic lattice with a density of $10.3 \mathrm{~g} \mathrm{~cm}^{-3}$ and the edge length of $314 \mathrm{pm}$ then, the structure of solid is
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simple cubic
Given, Density of element $(d)=10.3 \mathrm{~g} \mathrm{~cm}^{-3}$
Edge length $(a)=314 \mathrm{pm}$
Avogadro number $\left(N_A\right)=6.02 \times 10^{23} \mathrm{~mol}^{-1}$
Molar mass $(M)=247 \mathrm{~g} \mathrm{~mol}^{-1}$
$[\because Z=96$, having molar mass 247 , i.e. cesium (Cs) $]$
$\because \quad d=\frac{Z \times M}{a^3 \times N_A}$
where, $Z=$ number of lattice points in a unit cell
$\begin{aligned} & \therefore \quad \quad \quad Z=\frac{d \times a^3 \times N_A}{M} \\ & Z=\frac{10.3 \mathrm{~g} \mathrm{~cm}^{-3} \times(314)^3 \times 10^{-30} \mathrm{~cm} \times 6.02 \times 10^{23} \mathrm{~mol}^{-1}}{247 \mathrm{~g} \mathrm{~mol}^{-1}} \\ & Z=0.78 \approx 1\end{aligned}$
Thus, the structure of solid is simple cubic.
Edge length $(a)=314 \mathrm{pm}$
Avogadro number $\left(N_A\right)=6.02 \times 10^{23} \mathrm{~mol}^{-1}$
Molar mass $(M)=247 \mathrm{~g} \mathrm{~mol}^{-1}$
$[\because Z=96$, having molar mass 247 , i.e. cesium (Cs) $]$
$\because \quad d=\frac{Z \times M}{a^3 \times N_A}$
where, $Z=$ number of lattice points in a unit cell
$\begin{aligned} & \therefore \quad \quad \quad Z=\frac{d \times a^3 \times N_A}{M} \\ & Z=\frac{10.3 \mathrm{~g} \mathrm{~cm}^{-3} \times(314)^3 \times 10^{-30} \mathrm{~cm} \times 6.02 \times 10^{23} \mathrm{~mol}^{-1}}{247 \mathrm{~g} \mathrm{~mol}^{-1}} \\ & Z=0.78 \approx 1\end{aligned}$
Thus, the structure of solid is simple cubic.
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