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Current Question (ID: 8206)

Question:
At lower temperature, all gases except $\text{H}_2$ and He show:
Options:
  • 1. Negative deviation
  • 2. Positive deviation
  • 3. Positive and negative deviation
  • 4. None of the above
Solution:
Hint: The positive and negative deviation is determined with the help of the compressibility factor.\n\nExplanation:\n\n$\text{H}_2$ and He shows $PV>RT$; Rest all shows negative deviation.\n\nPositive deviation means volume increase, which is the result of a repulsive force, this indicates that it cannot liquified easily.\n\n-ve means, the volume got decreased due to attractive force, so becomes easier to liquify. Except for $\text{H}_2$, He, other gases show -ve deviations.

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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
}