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Current Question (ID: 8173)

Question:

\text{A mixture of He and Ar weighing } 5.0 \text{ g occupies a volume of } 10 \, \text{dm}^3 \text{ at } 27°\text{C} \text{ and } 1 \text{ atm pressure. The composition of the mixture is respectively -}

Options:

1. 70%, 30%
2. 25%, 75%
3. 75%, 25%
4. 40%, 60%

Solution:

Step 1: Let the mass of He be a g and the mass of Ar be b g in the mixture. Given the total mass: a + b = 5 ... (i) Step 2: Use the ideal gas equation to relate the moles of gases to the total pressure and volume. - Moles of He: n₁ = a/4 (molar mass of He = 4 g/mol) - Moles of Ar: n₂ = b/40 (molar mass of Ar = 40 g/mol) - Total moles: n₁ + n₂ Given: - Pressure (P) = 1 atm - Volume (V) = 10 L - Temperature (T) = 300 K (27°C) - Gas constant (R) = 0.0821 L atm K⁻¹ mol⁻¹ Substitute into the ideal gas equation: 1 × 10 = (a/4 + b/40) × 0.0821 × 300 Simplify: 10 = (10a + b)/40 × 24.63 10a + b = (10 × 40)/24.63 ≈ 16.24 ... (ii) Step 3: Solve equations (i) and (ii) to find a and b. From (i): b = 5 - a Substitute into (ii): 10a + (5 - a) = 16.24 9a + 5 = 16.24 9a = 11.24 a ≈ 1.25 g b = 5 - 1.25 = 3.75 g Step 4: Calculate the percentage composition. Percentage of He: (1.25/5) × 100 = 25% Percentage of Ar: (3.75/5) × 100 = 75% Hence, the composition of the mixture is 25% He and 75% Ar.

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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
}