Import Question JSON

Current Question (ID: 10550)

Question:

$\text{PH}_3 \text{ forms bubbles when passed slowly through water but } \text{NH}_3 \text{ gets dissolved in water, because:}$

Options:

1. $\text{NH}_3 \text{ forms an H-bond with water, so it is soluble but } \text{PH}_3 \text{ does not form an H-bond.}$ (Correct)
2. $\text{NH}_3 \text{ does not form an H-bond with water, so it is soluble but } \text{PH}_3 \text{ does form an H-bond.}$
3. $\text{NH}_3 \text{ and } \text{PH}_3 \text{ both form H-bond with water.}$
4. $\text{NH}_3 \text{ and } \text{PH}_3 \text{ both do not form H-bond with water.}$

Solution:

$\text{HINT: Nitrogen can form H-bond with water but P can't.}$ $\text{Explanation:}$ $\text{STEP 1: There are two requirements for hydrogen bonding.}$ $\text{(a) Molecule has hydrogen attached to a highly electronegative atom (N,O,F).}$ $\text{(b) Molecule has a lone pair of electrons on a small highly electronegative atom (N,O,F).}$ $\text{STEP 2:}$ $\text{The dissolution of } \text{NH}_3 \text{ and } \text{PH}_3 \text{ in water can be explained on the basis of H-bonding. } \text{NH}_3 \text{ forms H-bond with water so it is soluble but } \text{PH}_3 \text{ does not form H-bond with water so it remains as gas and forms bubble in water.}$

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
}