Import Question JSON

Current Question (ID: 8211)

Question:

The compressibility factor of a gas is less than unity at STP. The correct statement among the following is -

Options:

1. $V_m > 22.4$ L
2. $V_m < 22.4$ L
3. $V_m = 22.4$ L
4. $V_m = 44.8$ L

Solution:

Hint: $Z = \frac{PV}{nRT}$\nThe formula of compressibility factor of gas is as follows:\n\n$Z = \frac{PV}{nRT}$\n\nThe Z value is less than unity $(Z<1)$\n\n$1 > \frac{PV}{nRT}$\n\n$nRT > PV$\n\nThe value of nRT is 22.4 L.\n\nHence, P =1 atm, $22.4$ L $> V_M$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}