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Current Question (ID: 8182)

Question:

$\text{The volume of 0.0168 mol of } \text{O}_2 \text{ obtained by decomposition of } \text{KClO}_3 \text{ and collected by displacement of water is 428 ml at a pressure of 754 mm Hg at } 25^{\circ}\text{C}. \text{ The pressure of water vapour at } 25^{\circ}\text{C} \text{ is-}$

Options:

1. $18 \text{ mm Hg}$
2. $20 \text{ mm Hg}$
3. $22 \text{ mm Hg}$
4. $24 \text{ mm Hg}$

Solution:

$\text{Hint: } \text{P}_{\text{dry gas}} = \text{P}_{\text{total pressure}} - \text{P}_{\text{water vapour}}$\n\n$\text{Step 1:}$\n\n$\text{The container contains both } \text{O}_2 \text{ gas and water vapour. The total pressure is 754 mmHg.}$\n\n$\text{Calculate the pressure of water vapour as follows:}$\n\n$\text{P}_{\text{water vapour}} = \text{P}_{\text{total pressure}} - \text{P}_{\text{dry gas}}$\n\n$\text{Step 2:}$\n\n$\text{Calculate the pressure of } \text{O}_2 \text{ as follows:}$\n\n$P = \frac{nRT}{V}$\n\n$\text{The given values are as follows:}$\n\n$\text{n = 0.0168 mol}$\n\n$\text{R = 0.0821 atm L mol}^{-1}\text{K}^{-1}$\n\n$\text{T = }25^{\circ}\text{C = 298 K}$\n\n$\text{V = 428 mL = 0.428 L}$\n\n$P = \frac{0.0168 \times 0.0821 \times 298}{0.428}$\n\n$\text{P = 0.960 atm}$\n\n$\text{1 atm = 760 mmHg}$\n\n$= 0.960 \text{ atm} \times \frac{760 \text{ mmHg}}{1 \text{ atm}}$\n\n$\text{= 730 mmHg}$\n\n$\text{Step 3:}$\n\n$\text{P}_{\text{water vapour}} = 754 - 730$\n\n$\text{= 24 mmHg}$

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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
}