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Current Question (ID: 17961)

Question:

$\text{Copper crystallizes in a face-centered cubic lattice with a unit cell length of } 361 \text{ pm. The radius of the copper atom is:}$ $1. \ 128 \text{ pm}$ $2. \ 157 \text{ pm}$ $3. \ 181 \text{ pm}$ $4. \ 108 \text{ pm}$

Options:

1. $128 \text{ pm}$
2. $157 \text{ pm}$
3. $181 \text{ pm}$
4. $108 \text{ pm}$

Solution:

$\text{For FCC lattice, } r = \frac{\sqrt{2}a}{4}$ $\text{As given that copper crystallizes is FCC lattice (Face centred cubic). In FCC atoms are present on the corners of the cubic, unit cell as well as on the face centres of each face.}$ $\text{The atoms on the face diagonal will be touching each other. Let, the radius of the atom be } r \text{ and edge length of the cube be } a.$ $\text{Face diagonal of cube } = \sqrt{2}a$ $r + 2r + r = \sqrt{2}a$ $r = \frac{\sqrt{2}a}{4} = \frac{a}{2\sqrt{2}} = \frac{361}{2\sqrt{2}} \text{ pm} = 128 \text{ pm}$

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
}