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Current Question (ID: 8159)

Question:
The pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 $\text{dm}^3$ flask at 27$^\circ$C would be:
Options:
  • 1. $8.31 \times 10^4 \text{ Pa}$
  • 2. $7.43 \times 10^{-4} \text{ Pa}$
  • 3. $6.87 \times 10^3 \text{ Pa}$
  • 4. $8.31 \times 10^6 \text{ Pa}$
Solution:
Hint: Use Ideal Gas Equation\n\nStep 1: Calculate partial pressures\nFor methane ($\text{CH}_4$):\n$\text{P} = \frac{3.2 \text{ g} \times 8.314 \frac{\text{J}}{\text{mol K}} \times 300 \text{ K}}{16 \frac{\text{g}}{\text{mol}} \times 9 \text{ dm}^3 \times 0.01} = 55430 \text{ Pa}$\n\nFor carbon dioxide ($\text{CO}_2$):\n$\text{P} = \frac{4.4 \text{ g} \times 8.314 \frac{\text{J}}{\text{mol K}} \times 300 \text{ K}}{44 \frac{\text{g}}{\text{mol}} \times 9 \text{ dm}^3 \times 0.01} = 27710 \text{ Pa}$\n\nStep 2: Calculate total pressure\nTotal pressure $= 55430 \text{ Pa} + 27710 \text{ Pa} = 83140 \text{ Pa} \approx 8.31 \times 10^4 \text{ Pa}$\n\n(Note: 1 $\text{dm}^3$ = 0.001 $\text{m}^3$, temperature converted to Kelvin: 27$^\circ$C = 300 K)

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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
}