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Current Question (ID: 8168)

Question:

1.6 g \( O_2 \) gas and 0.1 g \( H_2 \) gas are placed in a 1.12 L flask at \( 0^\circ C \). The total pressure of the gas mixture will be-

Options:

1. 2 atm
2. 3 atm
3. 4 atm
4. 1 atm

Solution:

**Step 1:** Calculate the number of moles of each gas. - Moles of \( O_2 \): \[ n_{O_2} = \frac{\text{mass of } O_2}{\text{molar mass of } O_2} = \frac{1.6}{32} = 0.05 \text{ mol} \] - Moles of \( H_2 \): \[ n_{H_2} = \frac{\text{mass of } H_2}{\text{molar mass of } H_2} = \frac{0.1}{2} = 0.05 \text{ mol} \] **Step 2:** Calculate the total number of moles of the gas mixture. \[ n_t = n_{O_2} + n_{H_2} = 0.05 + 0.05 = 0.1 \text{ mol} \] **Step 3:** Use the ideal gas equation to find the total pressure. - Volume (\( V \)) = 1.12 L - Temperature (\( T \)) = \( 0^\circ C \) = 273 K - Gas constant (\( R \)) = 0.0821 L atm K\(^{-1}\) mol\(^{-1}\) \[ P_t = \frac{n_tRT}{V} = \frac{0.1 \times 0.0821 \times 273}{1.12} \approx 2 \text{ atm} \] Hence, the total pressure of the gas mixture is **2 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
}