MHT CET · Chemistry · Solid State
Calculate the molar mass of metal having density \(9.3 \mathrm{~g} \mathrm{~cm}^{-3}\) that forms simple cubic unit cell. \(\left[\mathrm{a}^3 \cdot \mathrm{N}_{\mathrm{A}}=22.6 \mathrm{~cm}^3 \mathrm{~mol}^{-1}\right]\)
- A \(210.2 \mathrm{~g} \mathrm{~mol}^{-1}\)
- B \(105.3 \mathrm{~g} \mathrm{~mol}^{-1}\)
- C \(52.6 \mathrm{~g} \mathrm{~mol}^{-1}\)
- D \(70.2 \mathrm{~g} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(210.2 \mathrm{~g} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
For simple cubic unit cell, \(\mathrm{n}=1\).
\(\text { Density }(\rho)=\frac{M n}{a^3 N_A}\)
\(\begin{aligned}
& 9.3=\frac{\mathrm{M} \times 1}{22.6} \\
& \mathrm{M}=\frac{9.3 \times 22.6}{1}=210.2 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}\)
\(\text { Density }(\rho)=\frac{M n}{a^3 N_A}\)
\(\begin{aligned}
& 9.3=\frac{\mathrm{M} \times 1}{22.6} \\
& \mathrm{M}=\frac{9.3 \times 22.6}{1}=210.2 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}\)
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