MHT CET · Chemistry · Solid State

Calculate the volume of unit cell of an element having molar mass \(27 \mathrm{~g} \mathrm{~mol}^{-1}\) that forms fce unit cell. \(\left[\rho . \mathrm{N}_{\mathrm{A}}=16.0 \times 10^{23} \mathrm{~g} \mathrm{~cm}^{-3} \mathrm{~mol}^{-1}\right]\)
[Note: The question has been modified to get the correct answer.]

A \(7.50 \times 10^{-23} \mathrm{~cm}^3\)
B \(6.75 \times 10^{-23} \mathrm{~cm}^3\)
C \(5.75 \times 10^{-23} \mathrm{~cm}^3\)
D \(8.25 \times 10^{-23} \mathrm{~cm}^3\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(6.75 \times 10^{-23} \mathrm{~cm}^3\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned}
& \rho=\frac{\mathrm{n} \times \mathrm{M}}{\mathrm{a}^3 \times \mathrm{N}_{\mathrm{A}}} \\
& \mathrm{a}^3=\frac{\mathrm{n} \times \mathrm{M}}{\rho \times \mathrm{N}_{\mathrm{A}}}=\frac{4 \times 27 \mathrm{~g} \mathrm{~mol}^{-1}}{16 \times 10^{23} \mathrm{~g} \mathrm{~cm}^{-3} \mathrm{~mol}^{-1}} \\
& \mathrm{~V}=\mathrm{a}^3=6.75 \times 10^{-23} \mathrm{~cm}^3
\end{aligned}\)
[Note: In the question, \(16.0 \times 10^{-23}\) is changed to \(16.0 \times 10^{23}\) to apply appropriate textual concepts.]

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From MHT CET

Explore more questions on app