MHT CET · Chemistry · Solid State
Calculate the molar mass of an element having density \(7.8 \mathrm{~g} \mathrm{~cm}^{-3}\) that forms bec unit cell. \(\left[\mathrm{a}^3 \cdot \mathrm{N}_{\mathrm{A}}=16.2 \mathrm{~cm}^3 \mathrm{~mol}^{-1}\right]\)
- A \(63.18 \mathrm{~g} \mathrm{~mol}^{-1}\)
- B \(61.23 \mathrm{~g} \mathrm{~mol}^{-1}\)
- C \(59.31 \mathrm{~g} \mathrm{~mol}^{-1}\)
- D \(65.61 \mathrm{~g} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(63.18 \mathrm{~g} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
For bec unit cell, \(\mathrm{n}=2\).
\(
\begin{aligned}
& \text {Density }(\rho)=\frac{\mathrm{Mn}}{\mathrm{a}^3 \mathrm{~N}_{\mathrm{A}}} \\
& 7.8=\frac{\mathrm{M} \times 2}{16.2} \\
& \mathrm{M}=\frac{7.8 \times 16.2}{2}=63.18 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
\)
\(
\begin{aligned}
& \text {Density }(\rho)=\frac{\mathrm{Mn}}{\mathrm{a}^3 \mathrm{~N}_{\mathrm{A}}} \\
& 7.8=\frac{\mathrm{M} \times 2}{16.2} \\
& \mathrm{M}=\frac{7.8 \times 16.2}{2}=63.18 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
\)
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