Solve this following

Question:

A diatomic molecule $X_{2}$ has a body-centred cubic (bcc) structure with a cell edge of $300 \mathrm{pm}$. The density of the molecule is $6.17 \mathrm{~g} \mathrm{~cm}^{-3}$. The number of molecules present in $200 \mathrm{~g}$ of $\mathrm{X}_{2}$ is

(Avogadro constant $\left(\mathrm{N}_{\mathrm{A}}\right)=6 \times 10^{23} \mathrm{~mol}^{-1}$ )

  1. $8 \mathrm{~N}_{\mathrm{A}}$

  2. $40 \mathrm{~N}_{\mathrm{A}}$

  3. $4 \mathrm{~N}_{\mathrm{A}}$

  4. $2 \mathrm{~N}_{\mathrm{A}}$


Correct Option: 3,

Solution:

$\mathrm{p}=\frac{2 \times \frac{\mathrm{M}}{\mathrm{N}_{\mathrm{A}}}}{\mathrm{a}^{3}} \Rightarrow 6.17=\frac{2 \times \frac{\mathrm{M}}{\mathrm{N}_{\mathrm{A}}}}{\left(3 \times 10^{-8} \mathrm{~cm}\right)^{3}}$

$\Rightarrow \mathrm{M} \simeq 50 \mathrm{gm} / \mathrm{mol}$

$\mathrm{No}=\frac{\mathrm{W}}{\mathrm{M}} \times \mathrm{N}_{\mathrm{A}}=\frac{200}{50} \times \mathrm{N}_{\mathrm{A}}=4 \mathrm{~N}_{\mathrm{A}}$

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