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Current Question (ID: 17955)

Question:

$\text{In a cubic unit cell, atom A is present at each corner, atom B at each face centre and atom C at the body centre. The simplest formula of the solid is:}$

Options:

1. $\text{AB}_2\text{C}$
2. $\text{A}_2\text{BC}$
3. $\text{AB}_3\text{C}$
4. $\text{ABC}_3$

Solution:

$\text{HINT: Cubic unit cell contains 8 corners and 6 faces.}$ $\text{Explanation:}$ $\text{An atom at the corner of a cube is shared among 8 unit cells. As there are 8 corners in a cube, a number of corner atoms (A) per unit cell = 8 \times \frac{1}{8} = 1.}$ $\text{A face-centered atom in a cube is shared by two unit cells. As there are 6 faces in a cube, a number of face-centered atoms (B) per unit cell = 6 \times \frac{1}{2} = 3.}$ $\text{An atom in the body of the cube is not shared by other cells.}$ \therefore \text{Number of atoms (C) at the body center per unit cell = 1}$ $\text{Hence, the formula of the solid is } \text{AB}_3\text{C}.$

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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
}