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Current Question (ID: 8185)

Question:

$\text{In a flask of volume V litres, 0.2 mol of oxygen, 0.4 mol of nitrogen, 0.1 mol of ammonia and 0.3 mol of helium are enclosed at } 27^{\circ}\text{C}. \text{ If the total pressure exerted by these non-reacting gases is one atmosphere, the partial pressure exerted by nitrogen is-}$

Options:

1. $1 \text{ atm}$
2. $0.1 \text{ atm}$
3. $0.2 \text{ atm}$
4. $0.4 \text{ atm}$

Solution:

$\text{Hint: Use Dalton Law of partial pressure formula}$\n\n$\text{Step 1:}$\n\n$\text{Dalton's Law of Partial Pressure states the total pressure exerted by a mixture of gases is equal to the sum of the partial pressure of each individual gas.}$\n\n$\text{P}_{\text{N}_2} = \text{Mol. fraction of } \text{N}_2 \times \text{P}_\text{T}$\n\n$\text{The total mole of gaseous mixture is 1.0 mol}$\n\n$\text{Step 2:}$\n\n$\text{Calculate the partial pressure of nitrogen as follows:}$\n\n$\text{P}_{\text{N}_2} = \text{Mol. fraction of } \text{N}_2 \times \text{Total pressure}$\n\n$= \frac{0.4}{0.2+0.4+0.1+0.3} \times 1 \text{ atm}$\n\n$= 0.4 \text{ atm}$

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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
}