MHT CET · Chemistry · Solid State
An element has a bcc structure with cell edge of \(288 \mathrm{pm}\). The density of element is \(7.2 \mathrm{~g} \mathrm{~cm}^{-3}\). What is the atomic mass of an element?
- A \(51.78\)
- B \(25.89\)
- C \(62.43\)
- D \(77.68\)
Answer & Solution
Correct Answer
(A) \(51.78\)
Step-by-step Solution
Detailed explanation
(D)
\(\mathrm{a}=288 \mathrm{pm}=2.88 \times 10^{-8} \mathrm{~cm} \)
\( \therefore \mathrm{a}^{3}=\left(2.88 \times 10^{-8}\right)^{3} \mathrm{~cm}^{3}=2.39\) \( \times ~10^{-23} \mathrm{~cm}^{3} \)
\( \rho=7.2 \mathrm{~g} \mathrm{~cm}^{-3}, \mathrm{n}=2(\text { for bcc cell }) \)
\( \therefore \mathrm{M}=\frac{\rho \times \mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}}{\mathrm{n}}=\frac{7.2 \times 2.39 \times 10^{-23} \times 6.022 \times 10^{23}}{2}\)\(=51.81 \mathrm{~g}\)
\(\mathrm{a}=288 \mathrm{pm}=2.88 \times 10^{-8} \mathrm{~cm} \)
\( \therefore \mathrm{a}^{3}=\left(2.88 \times 10^{-8}\right)^{3} \mathrm{~cm}^{3}=2.39\) \( \times ~10^{-23} \mathrm{~cm}^{3} \)
\( \rho=7.2 \mathrm{~g} \mathrm{~cm}^{-3}, \mathrm{n}=2(\text { for bcc cell }) \)
\( \therefore \mathrm{M}=\frac{\rho \times \mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}}{\mathrm{n}}=\frac{7.2 \times 2.39 \times 10^{-23} \times 6.022 \times 10^{23}}{2}\)\(=51.81 \mathrm{~g}\)
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