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Current Question (ID: 8106)

Question:

$\text{The incorrect order of boiling point is-}$

Options:

1. $\text{H}_2\text{O} > \text{CH}_3\text{OH}$
2. $\text{N}(\text{CH}_3)_3 < \text{NH}(\text{CH}_3)_2$
3. $\text{H}_3\text{PO}_4 > \text{Me}_3\text{PO}_4$
4. $\text{CH}_3\text{N}_3 > \text{HN}_3$

Solution:

$\text{Hint: The strength of hydrogen bonding increases, the boiling point also increases.}$ $\text{Water show stronger hydrogen bonding than alcohol hence, water has a higher boiling point than alcohol.}$ $\text{Tertiary amine did not show hydrogen bonding but secondary amine show hydrogen bonding hence, it has a higher boiling point than the tertiary amine.}$ $\text{The same thing is applicable in } \text{H}_3\text{PO}_4 \text{ and } \text{Me}_3\text{PO}_4\text{, } \text{H}_3\text{PO}_4 \text{ which show hydrogen bonding but } \text{Me}_3\text{PO}_4 \text{ does not show hydrogen bonding. Hence, the boiling point of } \text{H}_3\text{PO}_4 > \text{Me}_3\text{PO}_4\text{.}$ $\text{The correct order of B.P is}$ $\text{CH}_3\text{N}_3 < \text{HN}_3$ $\text{Among } \text{CH}_3\text{N}_3 \text{ there is dipole-dipole interaction while among } \text{HN}_3 \text{ intermolecular H-bonding occurs. Hence, } \text{HN}_3 \text{ has higher boiling point.}$

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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
}