Import Question JSON

Current Question (ID: 14713)

Question:

70 to 80 percent of electrolytes and water are reabsorbed in

Options:

1. Distal convoluted tubule
2. Proximal convoluted tubule
3. Ascending limb of Loop of Henle
4. Descending limb of Loop of Henle

Solution:

Maximum reabsorption of electrolytes and water occurs in PCT. The proximal convoluted tubule (PCT) is the first segment of the renal tubule and is responsible for the reabsorption of the majority of electrolytes and water from the glomerular filtrate. Approximately 70 to 80 percent of electrolytes and water are reabsorbed in the PCT. Option 1 (distal convoluted tubule) is incorrect because the distal convoluted tubule (DCT) is responsible for the reabsorption of some electrolytes, such as sodium and calcium ions, but it does not reabsorb the majority of electrolytes and water. Option 3 (ascending limb of the Loop of Henle) is incorrect because the ascending limb of the Loop of Henle is responsible for the reabsorption of sodium, potassium, and chloride ions, but it does not reabsorb the majority of electrolytes and water. Option 4 (descending limb of the Loop of Henle) is also incorrect because the descending limb of the Loop of Henle is responsible for the reabsorption of water, but it does not reabsorb the majority of electrolytes.

Import JSON File

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
}