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Current Question (ID: 11857)

Question:

In two Kingdom classification plantae had all bacteria, algae and fungi with them on the basis of

Options:

1. Presence of cell walls
2. Presence of nuclear membrane
3. Presence of some kind of body organization
4. Presence of different mode of nutrition

Solution:

In the two-kingdom classification system, plants (Plantae) included not only true plants (e.g., algae) but also bacteria and fungi. This classification was primarily based on the presence of cell walls, as bacteria, algae, and fungi all possess cell walls. However, this classification system has been largely replaced by more modern classification systems, such as the three-domain system (Archaea, Bacteria, and Eukarya) or the five-kingdom system (Monera, Protista, Fungi, Plantae, and Animalia), which take into account additional characteristics such as cellular structure, mode of nutrition, and genetic information.

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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
}