MHT CET · Chemistry · Solid State
Calculate the molar mass of an element having density \(5.6 \mathrm{~g} \mathrm{~cm}^{-3}\) that forms bcc structure. \(\left[\mathrm{a}^3 \times \mathrm{N}_{\mathrm{A}}=75 \mathrm{~cm}^3 \mathrm{~mol}^{-1}\right]\)
- A \(198 \mathrm{~g} \mathrm{~mol}^{-1}\)
- B \(210 \mathrm{~g} \mathrm{~mol}^{-1}\)
- C \(118 \mathrm{~g} \mathrm{~mol}^{-1}\)
- D \(225 \mathrm{~g} \mathrm{~mol}^{-1}\).
Answer & Solution
Correct Answer
(B) \(210 \mathrm{~g} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
For bcc unit cell, \(\mathrm{n}=2\).
\(\begin{aligned}
& \text { Density }(\rho)=\frac{M \times n}{a^3 \times N_A} \\
& 5.6 \mathrm{~g} \mathrm{~cm}^{-3}=\frac{M \times 2}{75 \mathrm{~cm}^3 \mathrm{~mol}^{-1}} \\
& M=\frac{5.6 \times 75}{2}=210 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}\)
\(\begin{aligned}
& \text { Density }(\rho)=\frac{M \times n}{a^3 \times N_A} \\
& 5.6 \mathrm{~g} \mathrm{~cm}^{-3}=\frac{M \times 2}{75 \mathrm{~cm}^3 \mathrm{~mol}^{-1}} \\
& M=\frac{5.6 \times 75}{2}=210 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}\)
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