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Current Question (ID: 18036)

Question:

$\text{Henry's law constant for the solution of methane in benzene at 298 K is } 4.27 \times 10^5 \text{ mm Hg. The mole fraction of methane in benzene at 298 K under 760 mm Hg will be:}$

Options:

1. $1.85 \times 10^{-5}$
2. $192 \times 10^{-4}$
3. $178 \times 10^{-5}$
4. $18.7 \times 10^{-5}$

Solution:

$\text{HINT: Use Henry's Law}$ $\text{Explanation:}$ $\text{STEP 1: Here, } p = 760 \text{ mm Hg}$ $k_H = 4.27 \times 10^5 \text{ mm Hg}$ $\text{STEP 2: According to Henry's law,}$ $p = k_H x$ $\Rightarrow x = \frac{p}{k_H}$ $\frac{760 \text{ mmHg}}{4.27 \times 10^5 \text{ mmHg}} = 177.99 \times 10^{-5} \approx 178 \times 10^{-5}$ $\text{Hence, the mole fraction of methane in benzene is } 178 \times 10^{-5}.$

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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
}