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Current Question (ID: 7419)

Question:

$\text{The vapour density of a mixture containing } \text{NO}_2 \text{ and } \text{N}_2\text{O}_4 \text{ is } 38.3. \text{ The mole of } \text{NO}_2 \text{ in a } 100 \text{ g mixture is:}$ $[\text{Vapour density} = (\text{Molar mass} / 2)]$

Options:

1. $0.437$
2. $0.347$
3. $0.557$
4. $0.663$

Solution:

$\text{Hint: Apply the mole concepts (equation based)}$ $\text{Given vapour density of mixture} = 38.3$ $\therefore \text{ Molar mass of mixture} = 2 \times 38.3 = 76.6$ $\text{Let a g of } \text{NO}_2 \text{ be present in } 100 \text{ g mixture,}$ $\therefore \text{ g of } \text{N}_2\text{O}_4 \text{ in mixture} = (100 - a)$ $\text{Therefore, } \frac{a}{46} + \frac{100-a}{92} = \frac{100}{76.6}$ $\frac{a}{46} + \frac{100-a}{92} = 1.30$ $2a + 100 - a = 1.30 \times 92$ $a + 100 = 120.10$ $\therefore a = 20.10 \text{ g}$ $\therefore \text{ Mole of } \text{NO}_2 \text{ in mixture} = 20.10/46$ $= 0.437$

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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
}