MHT CET · Chemistry · Solid State
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?
- A \(125 \cdot 5\)
- B \(100 \cdot 1\)
- C \(250 \cdot 0\)
- D \(150 \cdot 0\)
Answer & Solution
Correct Answer
(D) \(150 \cdot 0\)
Step-by-step Solution
Detailed explanation
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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