MHT CET · Chemistry · Solid State
Calculate the number of unit cells in 3 grams metal that crystallises to simple cubic unit cell having edge length \(336 \mathrm{pm}\). (density of metal \(=9.4 \mathrm{~g} \mathrm{~cm}^{-3}\) )
- A \(8.41 \times 10^{21}\)
- B \(6.25 \times 10^{21}\)
- C \(7.15 \times 10^{21}\)
- D \(5.82 \times 10^{21}\)
Answer & Solution
Correct Answer
(A) \(8.41 \times 10^{21}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{d}=\frac{\mathrm{Z} \times \mathrm{M}}{\mathrm{V} \times \mathrm{N}_{\mathrm{A}}}\)
\(\begin{aligned} & 9.4=\frac{1 \times M}{\left(3.36 \times 10^{-8}\right)^3 \times 6.02 \times 10^{23}} \\ & M=214.6 \mathrm{~g} / \mathrm{mol}\end{aligned}\)
\(\text {Number of moles atoms} =\frac{3}{214.6}\)
\(=0.01397\)
\(\text {Number of atoms} =0.01397 \times 6.02 \times 2 \times 10^{23}\)
\(=8.41 \times 10^{21}\)
\(\begin{aligned} & 9.4=\frac{1 \times M}{\left(3.36 \times 10^{-8}\right)^3 \times 6.02 \times 10^{23}} \\ & M=214.6 \mathrm{~g} / \mathrm{mol}\end{aligned}\)
\(\text {Number of moles atoms} =\frac{3}{214.6}\)
\(=0.01397\)
\(\text {Number of atoms} =0.01397 \times 6.02 \times 2 \times 10^{23}\)
\(=8.41 \times 10^{21}\)
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