MHT CET · Chemistry · Solid State
Calculate the volume of unit cell when metal having density \(1 \mathrm{~g} \mathrm{~cm}^{-3}\) and molar mass \(23 \mathrm{~g} \mathrm{~mol}^{-1}\) crystallises to form bce structure.
- A \(6.0 \times 10^{-23} \mathrm{~cm}^3\)
- B \(8.6 \times 10^{-23} \mathrm{~cm}^3\)
- C \(9.5 \times 10^{-23} \mathrm{~cm}^3\)
- D \(7.6 \times 10^{-23} \mathrm{~cm}^3\)
Answer & Solution
Correct Answer
(D) \(7.6 \times 10^{-23} \mathrm{~cm}^3\)
Step-by-step Solution
Detailed explanation
For \(b c c\) unit cell, \(\mathrm{n}=2\).
\(\operatorname{Density}(\rho)=\frac{M \times n}{a^3 \times N_A}\)
\(\text {Volume of unit cell }=\mathrm{a}^3=\frac{\mathrm{M} \times \mathrm{n}}{\rho \times \mathrm{N}_{\mathrm{A}}} \)
\( \)
\( \frac{23 \mathrm{~g} \mathrm{~mol}^{-1} \times 2}{1 \mathrm{~g} \mathrm{~cm}^{-3} \times 6.022 \times 10^{23} \mathrm{~mol}^{-1}} \)
\(=7.6 \times 10^{-23} \mathrm{~cm}^3\)
\(\operatorname{Density}(\rho)=\frac{M \times n}{a^3 \times N_A}\)
\(\text {Volume of unit cell }=\mathrm{a}^3=\frac{\mathrm{M} \times \mathrm{n}}{\rho \times \mathrm{N}_{\mathrm{A}}} \)
\( \)
\( \frac{23 \mathrm{~g} \mathrm{~mol}^{-1} \times 2}{1 \mathrm{~g} \mathrm{~cm}^{-3} \times 6.022 \times 10^{23} \mathrm{~mol}^{-1}} \)
\(=7.6 \times 10^{-23} \mathrm{~cm}^3\)
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