MHT CET · Chemistry · Solid State
An element with density \(2 \cdot 8 \mathrm{~g} \mathrm{~cm}^{-3}\) forms fec unit cell having edge length \(4 \times 10^{-8} \mathrm{~cm}\). Calculate molar mass of the element.
- A \(33 \cdot 0 \mathrm{~g} \mathrm{~mol}^{-1}\)
- B \(22 \cdot 0 \mathrm{~g} \mathrm{~mol}^{-1}\)
- C \(27 \cdot 0 \mathrm{~g} \mathrm{~mol}^{-1}\)
- D \(36 \cdot 0 \mathrm{~g} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(C) \(27 \cdot 0 \mathrm{~g} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\rho=2.8 \mathrm{gcm}^{-3}, \quad \mathrm{a}=4 \times 10^{-8} \mathrm{~cm}\),
\(\mathrm{n}=4(\) for fcc),\(\quad \mathrm{M}=?\)
\(\rho=\frac{\mathrm{n} \times \mathrm{M}}{\mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}} \quad \therefore \mathrm{M}=\frac{\rho \times \mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}}{\mathrm{n}}\)
\(\therefore M=\frac{2.8 \times\left(4 \times 10^{-8}\right)^{3} \times 6.022 \times 10^{23}}{4}\)\(=26.97 \approx 27 \mathrm{~g} \mathrm{~mol}^{-1}\)
\(\mathrm{n}=4(\) for fcc),\(\quad \mathrm{M}=?\)
\(\rho=\frac{\mathrm{n} \times \mathrm{M}}{\mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}} \quad \therefore \mathrm{M}=\frac{\rho \times \mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}}{\mathrm{n}}\)
\(\therefore M=\frac{2.8 \times\left(4 \times 10^{-8}\right)^{3} \times 6.022 \times 10^{23}}{4}\)\(=26.97 \approx 27 \mathrm{~g} \mathrm{~mol}^{-1}\)
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