Import Question JSON

Current Question (ID: 8191)

Question:

$\text{At STP, a container has 1 mole of He, 2 moles Ne, 3 moles of } \text{O}_2\text{, and 4 moles } \text{N}_2\text{. Without changing total pressure if 2 moles of } \text{O}_2 \text{ is removed, the partial pressure of } \text{O}_2 \text{ will be decreased by-}$

Options:

1. $26 \%$
2. $40 \%$
3. $58.33 \%$
4. $66.66 \%$

Solution:

$\text{Hint: The partial pressure of } \text{O}_2 \text{ will be decreased by}$\n\n$= \frac{\text{Change in } \text{O}_2 \text{ pressure}}{\text{Total Pressure of } \text{O}_2} \times 100$\n\n$\text{Step 1:}$\n\n$\text{Calculate the partial pressure of } \text{O}_2 \text{ as follows:}$\n\n$\text{P}_{\text{O}_2} = \text{P}_\text{T} \times \text{X}_{\text{O}_2}$\n\n$\text{P}_{\text{O}_2} = \text{P}_\text{T} \times \frac{\text{Number of mol of } \text{O}_2}{\text{Total mole(mole of } \text{O}_2\text{, Ne, He and } \text{N}_2\text{)}}$\n\n$\text{P}_{\text{O}_2} = \frac{3}{10} \times \text{P}_\text{T}$\n\n$\text{Step 2:}$\n\n$\text{After removing 2 moles of } \text{O}_2\text{, only one mole of } \text{O}_2 \text{ is left. The partial pressure of } \text{O}_2 \text{ is as follows:}$\n\n$\text{P}^{\prime}_{\text{O}_2} = \frac{1}{8} \times \text{P}_\text{T}$\n\n$\text{Decreasing in partial pressure of } \text{O}_2 = \frac{\text{Change in partial pressure of } \text{O}}{\text{Total pressure of } \text{O}_2}$\n\n$\text{Decreasing in partial pressure of } \text{O}_2 = \frac{\frac{3\text{P}_\text{T}}{10} - \frac{\text{P}_\text{T}}{8}}{\frac{3\text{P}_\text{T}}{10}} \times 100 = 58.33 \%$

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
}