MHT CET · Chemistry · Solid State
A metallic element crystallises in simple cubic lattice. If edge legth of the unit cell is \(3 A^{\circ}\), with density \(8 \mathrm{~g} / \mathrm{cc}\), what is the number of unit cells in \(100 \mathrm{~g}\) of the metal? \((\) Molar mass of metal \(=108 \mathrm{~g} / \mathrm{mol}\) )
- A \(1.33 \times 10^{20}\)
- B \(2 \times 10^{24}\)
- C \(2.7 \times 10^{22}\)
- D \(5 \times 10^{23}\)
Answer & Solution
Correct Answer
(D) \(5 \times 10^{23}\)
Step-by-step Solution
Detailed explanation
\((\mathrm{B})\)
\(\mathrm{a}=3 Å=3 \times 10^{-8} \mathrm{~cm}\)
Volume of the unit cell \(\left(\mathrm{a}^{3}\right)=\left(3 \times 10^{-8}\right)^{3} \mathrm{~cm}^{3}=27 \times 10^{-24} \mathrm{~cm}^{3}\)
Mass of unit cell \(=27 \times 10^{-24} \mathrm{~cm}^{3} \times 8 \mathrm{~g} \mathrm{~cm}^{-3}\)
\(=216 \times 10^{-24} \mathrm{~g}\)
\(\begin{aligned} 216 \times 10^{-24} \mathrm{~g}=1 \text { unit cell}\end{aligned}\)
\(\therefore 108 \mathrm{~g}=\frac{108}{216 \times 10^{-24}}=5 \times 10^{23} \text {unit cells.} \)
\(\mathrm{a}=3 Å=3 \times 10^{-8} \mathrm{~cm}\)
Volume of the unit cell \(\left(\mathrm{a}^{3}\right)=\left(3 \times 10^{-8}\right)^{3} \mathrm{~cm}^{3}=27 \times 10^{-24} \mathrm{~cm}^{3}\)
Mass of unit cell \(=27 \times 10^{-24} \mathrm{~cm}^{3} \times 8 \mathrm{~g} \mathrm{~cm}^{-3}\)
\(=216 \times 10^{-24} \mathrm{~g}\)
\(\begin{aligned} 216 \times 10^{-24} \mathrm{~g}=1 \text { unit cell}\end{aligned}\)
\(\therefore 108 \mathrm{~g}=\frac{108}{216 \times 10^{-24}}=5 \times 10^{23} \text {unit cells.} \)
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