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Current Question (ID: 8165)

Question:

If density of vapours of a substance of molar mass 18 g/mole at 1 atm pressure and 500 K is 0.36 kg m\(^{-3}\), then the value of Z for the vapours is- (R = 0.082 L atm mole K\(^{-1}\))

Options:

1. \(\frac{41}{50}\)
2. \(\frac{50}{41}\)
3. 1.1
4. 0.9

Solution:

**Step 1:** Calculate the real molar volume (\( V_{\text{real}} \)) using the given density. - Density (\( \rho \)) = 0.36 kg/m\(^3\) = 0.36 g/L (since 1 kg/m\(^3\) = 1 g/L) - Molar mass (\( M \)) = 18 g/mol \[ V_{\text{real}} = \frac{M}{\rho} = \frac{18}{0.36} = 50 \text{ L/mol} \] **Step 2:** Calculate the ideal molar volume (\( V_{\text{ideal}} \)) using the ideal gas equation. - Pressure (\( P \)) = 1 atm - Temperature (\( T \)) = 500 K - Gas constant (\( R \)) = 0.082 L atm K\(^{-1}\) mol\(^{-1}\) \[ V_{\text{ideal}} = \frac{RT}{P} = \frac{0.082 \times 500}{1} = 41 \text{ L/mol} \] **Step 3:** Calculate the compressibility factor (\( Z \)). \[ Z = \frac{V_{\text{real}}}{V_{\text{ideal}}} = \frac{50}{41} \] Hence, the value of \( Z \) for the vapours is \( \frac{50}{41} \).

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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
}