JEE Advanced · Chemistry · 15. Solid State
The density (in \(\mathrm{g} \mathrm{cm}^{-3}\) ) of the metal which forms a cubic close packed (ccp) lattice with an axial distance (edge length) equal to 400 pm is \(\qquad\) .
Use: Atomic mass of metal \(=105.6 \mathrm{amu}\) and Avogadro's constant \(=6 \times 10^{23} \mathrm{~mol}^{-1}\)
- A 11
- B 22
- C 33
- D 44
Answer & Solution
Correct Answer
(A) 11
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { CCP : } \\ & (\mathrm{Z}=4, \mathrm{a}=400 \mathrm{pm}, \mathrm{M}=105.6 \mathrm{~g} / \mathrm{mol}) \\ & \text { Density }=\frac{\mathrm{Z} \times \mathrm{M}}{\mathrm{a}^3 \times \mathrm{N}_{\mathrm{A}}} \\ & =\frac{4 \times 105.6}{\left(400 \times 10^{-10}\right)^3 \times 6 \times 10^{23}}=11.00 \mathrm{~g} / \mathrm{cm}^3\end{aligned}\)
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