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Current Question (ID: 8192)

Question:

$\text{The pressure of a 1 : 4 mixture of dihydrogen and dioxygen enclosed in a vessel is one atmosphere. The partial pressure of dioxygen is-}$

Options:

1. $0.8 \times 10^5 \text{ atm}$
2. $0.008 \text{ Nm}^{-2}$
3. $8 \times 10^4 \text{ Nm}^{-2}$
4. $0.25 \text{ atm}$

Solution:

$\text{Hint: Partial pressure of } \text{O}_2 = \text{Mole fraction of } \text{O}_2 \times \text{total pressure of mixture}$\n\n$\text{Step 1:}$\n\n$\text{The formula of partial pressure of } \text{O}_2 \text{ is as follows:}$\n\n$\text{Partial pressure of } \text{O}_2 = \text{Mole fraction of } \text{O}_2 \times \text{total pressure of mixture}$\n\n$\text{Step 2:}$\n\n$\text{The given values are as follows:}$\n\n$\text{The pressure of a 1 : 4 mixture of } \text{H}_2 \text{ and } \text{O}_2 \text{ enclosed in a vessel is one atmosphere. This suggests that the moles ratio of } \text{H}_2 \text{ and } \text{O}_2 \text{ is 1 : 4.}$\n\n$\text{Thus, partial pressure of dioxygen (} \text{O}_2 \text{) is given as}$\n\n$\therefore \text{ Partial pressure of } \text{O}_2 = \text{Mole fraction of } \text{O}_2 \times \text{total pressure of mixture}$\n\n$= \frac{4}{1+4} \times 1 = \frac{4}{5} \times 1 \text{ atm}$\n\n$= 0.8 \text{ atm} = 0.8 \times 10^5 \text{ Nm}^{-2} = 8 \times 10^4 \text{ Nm}^{-2}$

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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
}