MHT CET · Chemistry · Solid State
An element crystallises in a fcc lattice with cell edge \(250 \mathrm{pm}\). Calculate the density of an element
(at.mass \(=90 \cdot 3\) )
- A \(23 \cdot 12 \mathrm{~g} \mathrm{~cm}^{-3}\)
- B \(19 \cdot 20 \mathrm{~g} \mathrm{~cm}^{-3}\)
- C \(48 \cdot 40 \mathrm{~g} \mathrm{~cm}^{-3}\)
- D \(38.40 \mathrm{~g} \mathrm{~cm}^{-3}\)
Answer & Solution
Correct Answer
(D) \(38.40 \mathrm{~g} \mathrm{~cm}^{-3}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{a}=250 \mathrm{pm}=2.5 \times 10^{-8} \mathrm{~cm}\)
\(\mathrm{M}=90.3 \mathrm{~g} \mathrm{~mol}^{-1}, \mathrm{n}=4(\) for fcc cell \()\)
\(\rho=\frac{n \times M}{a^{3} \times N_{A}}=\frac{4 \times 90.3}{\left(2.5 \times 10^{-8}\right)^{3} \times 6.022 \times 10^{23}}\)
\(\therefore \rho=38.40 \mathrm{~g} \mathrm{~cm}^{-3}\)
\(\mathrm{M}=90.3 \mathrm{~g} \mathrm{~mol}^{-1}, \mathrm{n}=4(\) for fcc cell \()\)
\(\rho=\frac{n \times M}{a^{3} \times N_{A}}=\frac{4 \times 90.3}{\left(2.5 \times 10^{-8}\right)^{3} \times 6.022 \times 10^{23}}\)
\(\therefore \rho=38.40 \mathrm{~g} \mathrm{~cm}^{-3}\)
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