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Current Question (ID: 8162)

Question:

34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. The molar mass of phosphorus would be -

Options:

1. 1247.5 g mol\(^{-1}\)
2. 2234.7 g mol\(^{-1}\)
3. 3126.9 g mol\(^{-1}\)
4. 1134.6 g mol\(^{-1}\)

Solution:

**Step 1:** Given data: - Pressure (\( p \)) = 0.1 bar - Volume (\( V \)) = 34.05 mL = \( 34.05 \times 10^{-3} \) L - Temperature (\( T \)) = 546 °C = 819 K - Mass of phosphorus = 0.0625 g - Gas constant (\( R \)) = 0.083 bar L K\(^{-1}\) mol\(^{-1}\) **Step 2:** Calculate the number of moles (\( n \)) using the Ideal Gas Equation: \[ n = \frac{pV}{RT} \] \[ n = \frac{0.1 \times 34.05 \times 10^{-3}}{0.083 \times 819} \] \[ n \approx 5.07 \times 10^{-5} \text{ mol} \] **Step 3:** Calculate the molar mass (\( M \)) of phosphorus: \[ M = \frac{\text{given mass}}{\text{number of moles}} \] \[ M = \frac{0.0625}{5.07 \times 10^{-5}} \] \[ M \approx 1247.5 \text{ g mol}^{-1} \] Hence, the molar mass of phosphorus is \( 1247.5 \text{ g mol}^{-1} \).

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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
}