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Current Question (ID: 8164)

Question:

A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. The partial pressure of dihydrogen is -

Options:

1. 0.2 bar
2. 0.7 bar
3. 0.8 bar
4. 0.6 bar

Solution:

**Step 1:** Determine the number of moles of each gas. - Given: 20% by weight of dihydrogen (\( H_2 \)) and 80% dioxygen (\( O_2 \)). - Let the total mass of the mixture be 100 g. Then: - Mass of \( H_2 \) = 20 g - Mass of \( O_2 \) = 80 g - Molar mass of \( H_2 \) = 2 g/mol - Molar mass of \( O_2 \) = 32 g/mol - Moles of \( H_2 \) = \( \frac{20}{2} = 10 \) mol - Moles of \( O_2 \) = \( \frac{80}{32} = 2.5 \) mol **Step 2:** Calculate the mole fraction of \( H_2 \). - Total moles = \( 10 + 2.5 = 12.5 \) mol - Mole fraction of \( H_2 \) (\( x_{H_2} \)) = \( \frac{10}{12.5} = 0.8 \) **Step 3:** Determine the partial pressure of \( H_2 \). - Total pressure (\( P_{\text{total}} \)) = 1 bar - Partial pressure of \( H_2 \) (\( P_{H_2} \)) = \( x_{H_2} \times P_{\text{total}} = 0.8 \times 1 = 0.8 \) bar Hence, the partial pressure of dihydrogen is \( 0.8 \) bar.

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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
}