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Current Question (ID: 10956)

Question:
$\text{The most suitable method used for the separation of 1:1 mixture of ortho and para-nitrophenols is:}$
Options:
  • 1. $\text{Chromatography}$
  • 2. $\text{Crystallization}$
  • 3. $\text{Steam distillation}$ (Correct)
  • 4. $\text{Sublimation}$
Solution:
$\text{Hint: Intramolecular hydrogen bonding and intermolecular hydrogen bonding.}$ $\text{Steam distillation is used to purify the substances which}$ $\text{(i) are volatile in steam but are immiscible with water.}$ $\text{(ii) possess sufficiently high vapor pressure at the boiling point of water.}$ $\text{(iii) contain non-volatile impurities}$ $\text{Ortho = Intramolecular Hydrogen Bonding } \rightarrow \text{ Lowers the M.P/B.P}$ $\text{Para = Intermolecular Hydrogen Bonding } \rightarrow \text{ Increases the M.P/B.P}$ $\text{The boiling point of o-nitrophenol is lower than p-nitrophenol hence, o-nitrophenol separated out first as compare to p-nitrophenol.}$

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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
}