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Current Question (ID: 8146)

Question:
\text{120 g of an ideal gas of molecular weight 40 g/mol are confined to a volume of 20 L at 400 K. The pressure of the gas is} \\ (R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1})
Options:
  • 1. $ \text{3.90 atm} $
  • 2. $ \text{4.92 atm} $
  • 3. $ \text{6.02 atm} $
  • 4. $ \text{2.96 atm} $
Solution:
$ \text{Hint:} \n\n \text{Step 1:} \n\n \text{Calculate the moles of gas as follows:} \n\n \text{Number of mole} = \frac{\text{amount of compound}}{\text{molar mass of compound}} \n\n = \frac{120}{40} \n = 3 \text{ moles} \n\n \text{Step 2:} \n\n \text{Calculate the pressure of the gas is as follows:} \n\n P = \frac{nRT}{V} \n = \frac{3 \times 0.0821 \times 400}{20} \n = 4.92 \text{ atm.} $

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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
}