Import Question JSON

Current Question (ID: 17923)

Question:

$\text{The packing efficiency of the face-centered cubic unit cell is:}$ $\text{(assuming that the atoms are touching each other)}$

Options:

1. $52.40 \%$
2. $68.04 \%$
3. $74 \%$
4. $79.06 \%$

Solution:

$\text{HINT: The packing efficiency of FCC unit cell = 74\%}$ $\text{STEP 1: Draw the structure of FCC unit cell}$ $\text{Let the edge length of the unit cell be 'a' and the length of the face diagonal AC be b.}$ $\text{STEP 2:}$ $\text{Find out the relation between edge length and radius of the atom (r)}$ $\text{From } \triangle ABC, \text{ we have:}$ $AC^2 = BC^2 + AB^2$ $\Rightarrow b^2 = 2a^2$ $\Rightarrow b = a\sqrt{2}$ $\text{Also, from the figure; } b = 4r \Rightarrow a = 2r\sqrt{2}$ $\text{So, the volume of a cube, } a^3 = \left(2r\sqrt{2}\right)^3$ $\text{The volume occupied by unit cell = } 4 \times \frac{4}{3}\pi r^3 \text{ (as the number of atoms per unit cell = 4)}$ $\text{STEP 3:}$ $\text{Put all the values in the Packing efficiency formula.}$ $\text{Packing efficiency} = \frac{\text{Volume occupied by four spheres in the unit cell}}{\text{Total volume of the unit cell}} \times 100 \%$ $\Rightarrow \frac{16}{3}\pi r^3 \div 16\sqrt{2}r^3 \times 100 = 74\%$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}