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Current Question (ID: 17927)

Question:
$\text{The fraction of total volume occupied by the atoms present in a simple cube is:}$
Options:
  • 1. $\frac{\pi}{6}$
  • 2. $\frac{\pi}{3\sqrt{2}}$
  • 3. $\frac{\pi}{4\sqrt{2}}$
  • 4. $\frac{\pi}{4}$
Solution:
$\text{HINT: Packing fraction of Simple cubic crystal lattice is: } \frac{\pi}{6}$ $\text{Explanation:}$ $\text{STEP 1:}$ $\text{For simple cubic crystal:}$ $r = \frac{a}{2}$ $\text{where } r = \text{radius and } a \text{ is the edge length.}$ $\text{STEP 2:}$ $\text{Packing fraction} = \frac{\text{Volume occupied by atoms in the unit cell}}{\text{Volume of the unit cell}}$ $\text{Volume of the atom} = \frac{4}{3} \pi \left(\frac{a}{2}\right)^3$ $\text{Packing fraction} = \frac{\frac{4}{3} \pi \left(\frac{a}{2}\right)^3}{a^3} = \frac{\pi}{6}$

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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
}