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Current Question (ID: 13576)

Question:

Which one of the following phyla is correctly matched with its two general characteristics?

Options:

1. Arthropoda – Body divided into head, thorax and abdomen and respiration by mouth
2. Chordata – Notochord at early stage only and separate anal and urinary openings to the outside
3. Echinodermata – Pentamerous radial symmetry and mostly internal fertilization
4. Mollusca – Normally oviparous and development through a trochophore or veliger larva

Solution:

Mollusca is normally oviparous and its development may be direct or indirect. If development is indirect then it occurs through trochophore or veliger larva. Arthropoda is the largest phylum of animal kingdom. Body of arthropod is divisible into head, thorax and abdomen. Head and thorax often fused to form cephalothorax. The respiration takes place by general body surface, gills, tracheae or book lungs. Echinoderms have a pentamerous radial symmetry derived from an original bilateral symmetry. The fertilization is external, development indirect through free-swimming larval forms. Chordates are sharply distinguished from non-chordates by presence of notochord, dorsal tubular central nervous system and pharyngeal gills slits.

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}