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Current Question (ID: 8175)

Question:

A vessel of volume 0.02 m^{3} contains a mixture of hydrogen and helium at 20°C and 2 atm pressure. The mass of the mixture is 5 g. The ratio of the mass of hydrogen to that of helium in the mixture, is -

Options:

1. 1:2
2. 1:3
3. 2:3
4. 3:2

Solution:

**Step 1:** Calculate the total number of moles of the gaseous mixture using the ideal gas equation. - Volume (\( V \)) = 0.02 m^{3} = 20 L - Pressure (\( P \)) = 2 atm - Temperature (\( T \)) = 20°C = 293 K - Gas constant (\( R \)) = 0.0821 L atm K^{-1} mol^{-1} \[ n_{\text{total}} = \frac{PV}{RT} = \frac{2 \times 20}{0.0821 \times 293} \approx 1.67 \, \text{moles} \] **Step 2:** Let the mass of hydrogen (\( H_2 \)) be \( x \) g and the mass of helium (\( He \)) be \( 5 - x \) g. - Moles of \( H_2 \) = \( \frac{x}{2} \) (molar mass of \( H_2 \) = 2 g/mol) - Moles of \( He \) = \( \frac{5 - x}{4} \) (molar mass of \( He \) = 4 g/mol) Total moles: \[ \frac{x}{2} + \frac{5 - x}{4} = 1.67 \] Multiply through by 4 to eliminate denominators: \[ 2x + (5 - x) = 6.68 \] \[ x + 5 = 6.68 \] \[ x = 1.68 \, \text{g} \] Mass of helium: \[ 5 - x = 3.32 \, \text{g} \] **Step 3:** Compute the ratio of the mass of hydrogen to helium. \[ \text{Ratio} = \frac{1.68}{3.32} \approx \frac{1}{2} \] Hence, the ratio of the mass of hydrogen to that of helium in the mixture is **1:2**.

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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
}