MHT CET · Chemistry · Solid State
An element (molar mass 180) has BCC crystal structure with density \(18 \mathrm{~g} \mathrm{~cm}^{-3}\). What is the edge length of unit cell?
- A \(\sqrt[3]{23.2} \times 10^{-24} \mathrm{~cm}\)
- B \(\sqrt[3]{12.6} \times 10^{-24} \mathrm{~cm}\)
- C \(\sqrt[3]{33.2} \times 10^{-8} \mathrm{~cm}\)
- D \(\sqrt[3]{22.6} \times 10^{-8} \mathrm{~cm}\)
Answer & Solution
Correct Answer
(C) \(\sqrt[3]{33.2} \times 10^{-8} \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
\(
\mathrm{M}=180 \mathrm{~g} \mathrm{~mol}^{-1}, \rho=18 \mathrm{~g} \mathrm{~cm}^{-3} \text {, }
\)
For BCC crystal, \(\mathrm{z}=2, \mathrm{a}=\) ?
\(\rho=\frac{M \times z}{a^3 \times N_A} \therefore a^3=\frac{M \times z}{\rho \times N_A} \)
\( \therefore a^3=\frac{180 \mathrm{~g} \mathrm{~mol}^{-1} \times 2 \text { atom }}{18 \mathrm{~g} \mathrm{~cm}^{-3} \times 6.022 \times 10^{23} \text { atom mol }} \)
\( \therefore a^3=33.2 \times 10^{-24} \mathrm{~cm}^3 \)
\( \therefore a=\sqrt{33.2} \times 10^{-8} \mathrm{~cm}\)
\mathrm{M}=180 \mathrm{~g} \mathrm{~mol}^{-1}, \rho=18 \mathrm{~g} \mathrm{~cm}^{-3} \text {, }
\)
For BCC crystal, \(\mathrm{z}=2, \mathrm{a}=\) ?
\(\rho=\frac{M \times z}{a^3 \times N_A} \therefore a^3=\frac{M \times z}{\rho \times N_A} \)
\( \therefore a^3=\frac{180 \mathrm{~g} \mathrm{~mol}^{-1} \times 2 \text { atom }}{18 \mathrm{~g} \mathrm{~cm}^{-3} \times 6.022 \times 10^{23} \text { atom mol }} \)
\( \therefore a^3=33.2 \times 10^{-24} \mathrm{~cm}^3 \)
\( \therefore a=\sqrt{33.2} \times 10^{-8} \mathrm{~cm}\)
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