MHT CET · Chemistry · Solid State
Calculate molar mass of an element having density \(8.6 \mathrm{~g} \mathrm{~cm}^{-3}\) if it forms bcc structure \(\left[\mathrm{a}^3 \times \mathrm{N}_{\mathrm{A}}=22.0 \mathrm{~cm}^3 \mathrm{~mol}^{-1}\right]\)
- A \(106.18 \mathrm{~g} \mathrm{~mol}^{-1}\)
- B \(94.6 \mathrm{~g} \mathrm{~mol}^{-1}\)
- C \(88.25 \mathrm{~g} \mathrm{~mol}^{-1}\)
- D \(80.16 \mathrm{~g} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(B) \(94.6 \mathrm{~g} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
For bec unit cell, \(\mathrm{n}=2\).
\(\begin{aligned}
& \text { Density }(\rho)=\frac{\mathrm{M} \times \mathrm{n}}{\mathrm{a}^3 \times \mathrm{N}_{\mathrm{A}}} \\
& 8.6 \mathrm{~g} \mathrm{~cm}^{-3}=\frac{\mathrm{M} \times 2}{22 \mathrm{~cm}^3 \mathrm{~mol}^{-1}} \\
\therefore \quad & \mathrm{M}=\frac{8.6 \times 22}{2}=\frac{189.2}{2}=94.6 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}\)
\(\begin{aligned}
& \text { Density }(\rho)=\frac{\mathrm{M} \times \mathrm{n}}{\mathrm{a}^3 \times \mathrm{N}_{\mathrm{A}}} \\
& 8.6 \mathrm{~g} \mathrm{~cm}^{-3}=\frac{\mathrm{M} \times 2}{22 \mathrm{~cm}^3 \mathrm{~mol}^{-1}} \\
\therefore \quad & \mathrm{M}=\frac{8.6 \times 22}{2}=\frac{189.2}{2}=94.6 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}\)
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