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Current Question (ID: 17941)

Question:
$\text{A solid has a bcc structure. If the distance of closest approach between the two atoms is } 1.73 \, \text{Å, then the edge length of the cell is:}$
Options:
  • 1. $200 \, \text{pm}$
  • 2. $\frac{\sqrt{3}}{\sqrt{2}} \, \text{pm}$
  • 3. $142.2 \, \text{pm}$
  • 4. $\sqrt{2} \, \text{pm}$
Solution:
$\text{HINT: For bcc lattice, closest approach is } \frac{1}{2} \text{ of body diagonal.}$ $\text{Explanation:}$ $r_{\text{atom}} = \frac{\sqrt{3}}{4}a; \text{ Also closest approach in bcc lattice is } \frac{1}{2} \text{ of body diagonal,}$ $\text{i.e, } \frac{\sqrt{3}}{4}a = 1.73 \, \text{Å}$ $\text{or } a = 1.73 \times 2 / \sqrt{3} = 1.996 \, \text{Å} = 199.6 \, \text{pm}$ $\text{Distance of closest approach between two atoms } = \frac{\sqrt{3}}{2} = 1.73 \, \text{(given)}$ $\text{Let the side length be } a.$ $l = \frac{a \sqrt{3}}{2} = \text{Distance of closest approach}$ $\frac{a \sqrt{3}}{2} = \sqrt{3} \times 1.73 \, \text{Å}$ $a = 2 \times 1.73 \, \text{Å}$ $a = 200 \, \text{pm}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}