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Current Question (ID: 10714)

Question:
$\text{Which of the following gases is the easiest to liquefy?}$
Options:
  • 1. $\text{Ar}$
  • 2. $\text{Ne}$
  • 3. $\text{Xe}$ (Correct)
  • 4. $\text{Kr}$
Solution:
$\text{HINT: Xe is easily liquefiable gas.}$ $\text{Explanation:}$ $\text{The critical temperature signifies the force of attraction between the molecules. The higher the critical temperature, higher is the intermolecular force of attraction and easier is the liquefaction of the gas. Xe has the highest critical temperature among all the given option. Hence, Xe is the correct answer.}$ $\text{The critical temperature of noble gas is as follows:}$ $\text{He} = -267.96 \, ^\circ\text{C}$ $\text{Ne} = -228.75 \, ^\circ\text{C}$ $\text{Ar} = -122.4 \, ^\circ\text{C}$ $\text{Kr} = -63.8 \, ^\circ\text{C}$ $\text{Xe} = 16.6 \, ^\circ\text{C}$ $\text{Rn} = 103 \, ^\circ\text{C}$

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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
}