Import Question JSON

Current Question (ID: 8103)

Question:
$\text{The type of molecular forces of attraction present in the following compound is:}$
Options:
  • 1. $\text{Intermolecular H-bonding.}$
  • 2. $\text{Intramolecular H-bonding.}$
  • 3. $\text{Van der Waals force.}$
  • 4. $\text{All of the above.}$
Solution:
$\text{HINT: Large non bonded molecules have van der Waals forces.}$ $\text{Explanation:}$ $\text{The given compound shows all three types of a force of interaction. It has intramolecular hydrogen bonding, it also shows intermolecular hydrogen bonding with other molecules. Due to the benzene ring, it also shows Van der Waals type of force of attraction with other molecules.}$ $\text{The compound exhibits:}$ $\text{1. Intramolecular H-bonding: Between the OH group and NO}_2 \text{ group within the same molecule}$ $\text{2. Intermolecular H-bonding: Between OH groups of different molecules}$ $\text{3. Van der Waals forces: Due to the aromatic benzene ring system}$

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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
}