Import Question JSON

Current Question (ID: 8190)

Question:
$\text{In water-saturated air, the mole fraction of water vapor is 0.02. If the total pressure of the saturated air is 1.2 atm, the partial pressure of dry air is-}$
Options:
  • 1. $1.17 \text{ atm}$
  • 2. $1.76 \text{ atm}$
  • 3. $1.27 \text{ atm}$
  • 4. $0.98 \text{ atm}$
Solution:
$\text{Hint: } \text{P}_{\text{total}} = \text{P}_{\text{dry air}} + \text{P}_{\text{H}_2\text{O}}$\n\n$\text{Step 1:}$\n\n$\text{The partial pressure of dry gas = Total pressure x mole fraction of dry air}$\n\n$\text{P}_{\text{H}_2\text{O}} = \chi_{\text{H}_2\text{O}}\text{P}_{\text{total}}$\n\n$= 0.02 \times 1.2$\n\n$= \frac{24}{1000} = 0.024 \text{ atm}$\n\n$\text{Step 2:}$\n\n$\text{P}_{\text{total}} = \text{P}_{\text{dry air}} + \text{P}_{\text{H}_2\text{O}}$\n\n$\text{P}_{\text{dry air}} = \text{P}_{\text{total}} - \text{P}_{\text{H}_2\text{O}}$\n\n$= \text{P}_{\text{total}} - \text{P}_{\text{H}_2\text{O}}$\n\n$= 1.2 - 0.024$\n\n$= 1.176 \text{ atm}$

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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
}