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Current Question (ID: 17964)

Question:
$\text{A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy } \frac{1}{3} \text{rd of tetrahedral voids. The formula of the compound would be -}$
Options:
  • 1. $\text{MN}$
  • 2. $\text{M}_3\text{N}_4$
  • 3. $\text{M}_2\text{N}_3$
  • 4. $\text{M}_3\text{N}_2$
Solution:
$\text{HINT: Number of tetrahedral voids} = 2 \times \text{(Number of octahedral voids)}$ $\text{STEP 1:}$ $\text{The ccp lattice is formed by the atoms of the element N. In ccp, atoms are present at the corners and at the face center.}$ $\text{Number of N atoms in a unit cell} = 4$ $\text{The total number of tetrahedral voids} = 8$ $\text{STEP 2:}$ $\text{The atoms of element M occupy} \frac{1}{3} \text{rd of the tetrahedral voids.}$ $\text{Therefore, the number of atoms of M} = 8 \times \frac{1}{3} = \frac{8}{3} \text{rd of the number of atoms of N.}$ $\text{Therefore, the ratio of the number of atoms of M to that of N is M:N}$ $= \frac{8}{3} : 4 = 8 : 12 \approx 2 : 3$ $\text{Thus, the formula of the compound is } \text{M}_2\text{N}_3.$

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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
}