Q42 / 150 · MHT CET 2024 (03 May Shift 2) · Chemistry

Calculate the volume of unit cell of an element having molar mass $63.5 \mathrm{~g} \mathrm{~mol}^{-1}$. that forms fce structure $\left[\varrho \times \mathrm{N}_{\mathrm{A}}=5.5 \times 10^{24} \mathrm{~g} \mathrm{~cm}^{-3} \mathrm{~mol}^{-1}\right]$

  • A
    $4.1 .02 \times 10^{-25} \mathrm{~cm}^3$
  • B
    $5.430 \times 10^{-23} \mathrm{~cm}^3$
  • C
    $5.014 \times 10^{-23} \mathrm{~cm}^3$
  • Correct answer
    D
    $4.618 \times 10^{-23} \mathrm{~cm}^3$

Solution

$\begin{aligned} & \rho=\frac{\mathrm{n} \times \mathrm{M}}{\mathrm{a}^3 \mathrm{~N}_{\mathrm{A}}} \\ & \text { volume of unit cell }=\mathrm{a}^3=\frac{\mathrm{n} \times \mathrm{M}}{\rho \times \mathrm{N}_{\mathrm{A}}} \\ & =\frac{4 \times 63.5 \mathrm{~g} \mathrm{~mol}^{-1}}{5.5 \times 10^{24} \mathrm{~g} \mathrm{~cm}^{-3} \mathrm{~mol}^{-1}}=4.618 \times 10^{-23} \mathrm{~cm}^3\end{aligned}$

Hence, the correct answer is option: 'D'.

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