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An element crystallises bcc type of unit cell, the density and edge length of unit
cell is $4 \mathrm{~g} \mathrm{~cm}^{-3}$ and 500 pm respectively. What is the atomic mass of an element?
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cell is $4 \mathrm{~g} \mathrm{~cm}^{-3}$ and 500 pm respectively. What is the atomic mass of an element?
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The correct answer is:
$150 \cdot 0$
In bcc, $\mathrm{n}=2, \rho=4 \mathrm{~g} \mathrm{~cm}^{-3}$
$\mathrm{a}=500 \mathrm{pm}=500 \times 10^{-10} \mathrm{~cm}=5 \times 10^{-8} \mathrm{~cm}$
$\therefore$ Volume of unit cell, $\mathrm{a}^{3}=\left(5 \times 10^{-8} \mathrm{~cm}\right)^{3}=125 \times 10^{-24} \mathrm{~cm}^{3}$
Now, $\mathrm{M}=\frac{\rho \times \mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}}{\mathrm{n}}=\frac{4 \times 125 \times 10^{-24} \times 6.022 \times 10^{23}}{2}=150.5 \mathrm{~g} \mathrm{~mol}^{-1}$
$\mathrm{a}=500 \mathrm{pm}=500 \times 10^{-10} \mathrm{~cm}=5 \times 10^{-8} \mathrm{~cm}$
$\therefore$ Volume of unit cell, $\mathrm{a}^{3}=\left(5 \times 10^{-8} \mathrm{~cm}\right)^{3}=125 \times 10^{-24} \mathrm{~cm}^{3}$
Now, $\mathrm{M}=\frac{\rho \times \mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}}{\mathrm{n}}=\frac{4 \times 125 \times 10^{-24} \times 6.022 \times 10^{23}}{2}=150.5 \mathrm{~g} \mathrm{~mol}^{-1}$
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