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Current Question (ID: 18410)

Question:

$\text{In an adsorption experiment, a graph between } \log(x/m) \text{ versus } \log P \text{ was found to be linear with a slope of } 45^\circ. \text{ The intercept on the y-axis was found to be } 0.3. \text{ What will be the value of } x/m \text{ at a pressure of } 3 \text{ atm for the Freundlich adsorption isotherm when } 1/n \text{ remains constant?}$ $[\text{Antilog } 0.3 = 2]$

Options:

1. $2$
2. $4$
3. $6$
4. $5$

Solution:

$\text{HINT: } \frac{x}{m} = kP^{1/n}$ $\text{Explanation:}$ $\text{By using equation of Freundlich isotherm, we get}$ $\log \frac{x}{m} = \log k + \frac{1}{n} \log P$ $\log k = 0.3 \Rightarrow k = 2$ $\frac{x}{m} = kP^{1/n}$ $\frac{x}{m} = 2 \times 3 = 6 \quad [1/n \text{ remains constant}]$

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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
}