MHT CET · Chemistry · Solid State
Calculate the volume of unit cell if an element having molar mass \(92 \mathrm{~g} \mathrm{~mol}^{-1}\) that forms bcc structure \(\left[\varrho \times \mathrm{N}_{\mathrm{A}}=5.0 \times 10^{24} \mathrm{~g} \mathrm{~cm}^{-3} \mathrm{~mol}^{-1}\right]\)
- A \(2.44 \times 10^{-23} \mathrm{~cm}^3\)
- B \(5.86 \times 10^{-23} \mathrm{~cm}^3\)
- C \(3.68 \times 10^{-23} \mathrm{~cm}^3\)
- D \(4.76 \times 10^{-23} \mathrm{~cm}^3\).
Answer & Solution
Correct Answer
(C) \(3.68 \times 10^{-23} \mathrm{~cm}^3\)
Step-by-step Solution
Detailed explanation
For bcc unit cell, \(\mathrm{n}=2\).
\(\text {Density of bec unit cell }=\rho=\frac{M \times n}{a^3 \times N_A}\)
\(\therefore \text {Volume of unit cell }\left(\mathrm{a}^3\right) =\frac{\mathrm{M} \times \mathrm{n}}{\rho \times \mathrm{N}_{\mathrm{A}}} \)
\( =\frac{92 \mathrm{~g} \mathrm{~mol}^{-1} \times 2}{5 \times 10^{24} \mathrm{~g} \mathrm{~cm}^{-3} \mathrm{~mol}^{-1}} \)
\( =3.68 \times 10^{-23} \mathrm{~cm}^3\)
\(\text {Density of bec unit cell }=\rho=\frac{M \times n}{a^3 \times N_A}\)
\(\therefore \text {Volume of unit cell }\left(\mathrm{a}^3\right) =\frac{\mathrm{M} \times \mathrm{n}}{\rho \times \mathrm{N}_{\mathrm{A}}} \)
\( =\frac{92 \mathrm{~g} \mathrm{~mol}^{-1} \times 2}{5 \times 10^{24} \mathrm{~g} \mathrm{~cm}^{-3} \mathrm{~mol}^{-1}} \)
\( =3.68 \times 10^{-23} \mathrm{~cm}^3\)
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