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Current Question (ID: 18404)

Question:

$\text{In the Freundlich adsorption isotherm, the value of } 1/n \text{ is:}$ $1. \text{ Between 0 and 1 in all cases.}$ $2. \text{ Between 2 and 4 in all cases.}$ $3. \text{ 1 in the case of physical adsorption.}$ $4. \text{ 1 in the case of chemisorption.}$

Options:

1. $\text{Between 0 and 1 in all cases.}$
2. $\text{Between 2 and 4 in all cases.}$
3. $\text{1 in the case of physical adsorption.}$
4. $\text{1 in the case of chemisorption.}$

Solution:

$\text{HINT: } \frac{x}{m} = kp^{1/n}$ $\text{Explanation:}$ $\text{According to the Freundlich adsorption isotherm,}$ $\frac{x}{m} = kp^{1/n}$ $n = \text{ always greater than 1.}$ $\text{Therefore, the value of } 1/n \text{ lies between 0 and 1 in all cases.}$

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
}