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Current Question (ID: 8177)

Question:

\text{A collapsed polythene bag of 30 litre capacity is partially blown up by the addition of 10 litre of } N_2 \text{ at 0.965 atm at 298 K. Subsequently, enough } O_2 \text{ is pumped into the bag so that at 298 K and external pressure of 0.990 atm, the bag contains full 30 litre. The final pressure of } O_2 \text{ in the following experiment is-}

Options:

1. 0.67 atm
2. 0.52 atm
3. 85 atm
4. 20 atm

Solution:

\textbf{Step 1:} \text{ Calculate the partial pressure of } N_2 \text{ after expansion to 30 L.} \text{- Initial volume of } N_2 \text{ (} V_1 \text{) = 10 L} \text{- Initial pressure of } N_2 \text{ (} P_1 \text{) = 0.965 atm} \text{- Final volume of } N_2 \text{ (} V_2 \text{) = 30 L} \text{- Temperature remains constant at 298 K.} \text{Using Boyle's Law:} P_1 V_1 = P_2 V_2 0.965 \times 10 = P_2 \times 30 P_2 = \frac{0.965 \times 10}{30} \approx 0.32 \text{ atm} \textbf{Step 2:} \text{ Determine the partial pressure of } O_2 \text{ to achieve the total pressure.} \text{- Total pressure (} P_T \text{) = 0.990 atm} \text{- Partial pressure of } N_2 \text{ (} P_{N_2} \text{) = 0.32 atm} P_T = P_{N_2} + P_{O_2} 0.990 = 0.32 + P_{O_2} P_{O_2} = 0.990 - 0.32 = 0.67 \text{ atm} \text{Hence, the final pressure of } O_2 \text{ is } \textbf{0.67 atm}\text{.}

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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
}