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Current Question (ID: 8160)

Question:

\text{Equal masses of } H_2\text{, } O_2 \text{ and methane have been taken in a container of volume V at temperature 27°C in identical conditions. The ratio of the volumes of gases } H_2:O_2:CH_4 \text{ would be-}

Options:

1. \text{The ratio is } 8:16:1
2. \text{The ratio is } 16:8:1
3. \text{The ratio is } 16:1:2
4. \text{The ratio is } 8:1:2

Solution:

\text{According to Avogadro's hypothesis: Volume } V \text{ is proportional to moles } n\\ \\ \text{Volume ratio } = \text{ Mole ratio}\\ V_{H_2} : V_{O_2} : V_{CH_4} = n_{H_2} : n_{O_2} : n_{CH_4}\\ \\ \text{Since moles } = \text{ mass/molecular weight:}\\ n = \frac{m}{M}\\ \\ \text{With equal masses:}\\ \frac{m}{2} : \frac{m}{32} : \frac{m}{16} = 16 : 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
}