MHT CET · Chemistry · Solid State

Calculate the number of unit cells in \(38.6 \mathrm{~g}\) of noble metal haven density \(19 \cdot 3 \mathrm{~g} \mathrm{~cm}^{-3}\) and volume of one unit cell is \(6 \cdot 18 \times 10^{-23} \mathrm{~cm}^{3}\) ?

A \(3 \cdot 236 \times 10^{22}\)
B \(6 \cdot 180 \times 10^{23}\)
C \(6 \cdot 236 \times 10^{20}\)
D \(3 \cdot 236 \times 10^{23}\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(3 \cdot 236 \times 10^{22}\)

Step-by-step Solution

Detailed explanation

\(\text {Vol. of metal }=\frac{\text { Mass }}{\text { Density }}=\frac{38.6 \mathrm{~g}}{19.3 \mathrm{~g} \mathrm{~cm}^{-3}}\) \( =2 \mathrm{~cm}^{3} \)
\( \text {No. of unit cells in } 38.6 \mathrm{~g} \text { of noble metal } =\) \(\frac{\text {Total vol. of metal }}{\text {Vol. of one unit cell }} \)
\( =\frac{2 \mathrm{~cm}^{3}}{6.18 \times 10^{-23} \mathrm{~cm}^{3}}=3.236 \times 10^{22}\)

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.

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