MHT CET · Chemistry · Solid State
Calculate the number of atoms in 20 gram metal which crystalises to simple cubic structure having unit cell length \(340 \mathrm{pm}\). (density of metal \(=9.8 \mathrm{~g} \mathrm{~cm}^{-3}\) )
- A \(5.81 \times 10^{22}\)
- B \(5.19 \times 10^{22}\)
- C \(5.42 \times 10^{22}\)
- D \(4.95 \times 10^{22}\)
Answer & Solution
Correct Answer
(B) \(5.19 \times 10^{22}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{d}=\frac{\mathrm{Z} \times \mathrm{M}}{\mathrm{V} \times \mathrm{N}_{\mathrm{A}}}\)
\(9.8=\frac{1 \times M}{\left(3.40 \times 10^{-8}\right)^3 \times 6.02 \times 10^{23}}\)
\(\begin{aligned} & M=\frac{9.8 \times 6.02 \times 39.3}{10} \\ & =231.87\end{aligned}\)
Moles of atom in \(20 \mathrm{~g}\) sample
\(=\frac{20}{231.87}=0.08625\)
No. of atoms \(=0.08625 \times 6.02 \times 10^{23}\)
\(=5.19 \times 10^{22}\)
\(9.8=\frac{1 \times M}{\left(3.40 \times 10^{-8}\right)^3 \times 6.02 \times 10^{23}}\)
\(\begin{aligned} & M=\frac{9.8 \times 6.02 \times 39.3}{10} \\ & =231.87\end{aligned}\)
Moles of atom in \(20 \mathrm{~g}\) sample
\(=\frac{20}{231.87}=0.08625\)
No. of atoms \(=0.08625 \times 6.02 \times 10^{23}\)
\(=5.19 \times 10^{22}\)
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