Import Question JSON

Current Question (ID: 17921)

Question:

$\text{The percentage of free space in a body-centered cubic unit cell is:}$

Options:

1. $30\%$
2. $32\%$
3. $34\%$
4. $28\%$

Solution:

$\text{STEP 1: The packing fraction can be calculated using the following formula:}$ $\text{Packing fraction} = \frac{\text{Volume occupied by atoms in a unit cell}}{\text{Volume of the unit cell}}$ $\text{STEP 2: Calculate the volume occupied by atoms in a unit cell and the volume of a unit cell as follows:}$ $\text{For body-centered cube, number of atoms} = 2$ $\text{The volume occupied by two atoms} = 2 \times \frac{4}{3} \pi r^3$ $\text{The volume of a unit cell (cube)} = a^3$ $\text{For body-centered cubic unit cell, a value is} = \frac{4}{\sqrt{3}} r$ $\text{The volume of a unit cell} = \left(\frac{4}{\sqrt{3}} r\right)^3$ $\text{STEP 3: Calculate the packing fraction as follows:}$ $\text{Packing fraction} = \frac{\text{Volume occupied by atoms in a unit cell}}{\text{Volume of the unit cell}}$ $\text{Packing fraction} = \frac{2 \times \frac{4}{3} \pi r^3}{\left(\frac{4}{\sqrt{3}} r\right)^3} = \frac{\sqrt{3} \pi}{8} = 0.68$ $\text{STEP 4:}$ $\text{The volume occupied by atoms in a unit cell} = 68\%$ $\text{and volume unoccupied or free space} = 100 - 68$ $= 32\%$ $\text{Hence, option 2 is the correct answer.}$

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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
}