MHT CET · Chemistry · Solid State
Calculate the volume of unit cell of an element having molar mass \(63.5 \mathrm{~g} \mathrm{~mol}^{-1}\). that forms fce structure \(\left[\varrho \times \mathrm{N}_{\mathrm{A}}=5.5 \times 10^{24} \mathrm{~g} \mathrm{~cm}^{-3} \mathrm{~mol}^{-1}\right]\)
- A \(4.1 .02 \times 10^{-25} \mathrm{~cm}^3\)
- B \(5.430 \times 10^{-23} \mathrm{~cm}^3\)
- C \(5.014 \times 10^{-23} \mathrm{~cm}^3\)
- D \(4.618 \times 10^{-23} \mathrm{~cm}^3\)
Answer & Solution
Correct Answer
(D) \(4.618 \times 10^{-23} \mathrm{~cm}^3\)
Step-by-step Solution
Detailed explanation
\(\rho=\frac{\mathrm{n} \times \mathrm{M}}{\mathrm{a}^3 \mathrm{~N}_{\mathrm{A}}} \)
\( \text { volume of unit cell }=\mathrm{a}^3=\frac{\mathrm{n} \times \mathrm{M}}{\rho \times \mathrm{N}_{\mathrm{A}}} \)
\( =\frac{4 \times 63.5 \mathrm{~g} \mathrm{~mol}^{-1}}{5.5 \times 10^{24} \mathrm{~g} \mathrm{~cm}^{-3} \mathrm{~mol}^{-1}}=4.618 \times 10^{-23} \mathrm{~cm}^3\)
\( \text { volume of unit cell }=\mathrm{a}^3=\frac{\mathrm{n} \times \mathrm{M}}{\rho \times \mathrm{N}_{\mathrm{A}}} \)
\( =\frac{4 \times 63.5 \mathrm{~g} \mathrm{~mol}^{-1}}{5.5 \times 10^{24} \mathrm{~g} \mathrm{~cm}^{-3} \mathrm{~mol}^{-1}}=4.618 \times 10^{-23} \mathrm{~cm}^3\)
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