Import Question JSON

Current Question (ID: 18455)

Question:

$\text{Hydrophobic sols are easily coagulated as?}$

Options:

1. $\text{The particles present in them come closer and form aggregates, leading to precipitation.}$
2. $\text{The particles present in them move farther and form aggregates, which leads to precipitation.}$
3. $\text{The particles present in them come closer and do not form aggregates, leading to precipitation.}$
4. $\text{The particles present in them move farther and do not form aggregates, leading to precipitation.}$

Solution:

$\text{Lyophobic sols are irreversible in nature.}$ $\text{When substances such as metals and their sulphides etc. are mixed with the dispersion medium, they do not form colloidal sols.}$ $\text{Their colloidal sols can be prepared only by special methods.}$ $\text{Such sols are called lyophobic sols.}$ $\text{These sols are irreversible in nature.}$ $\text{For example: sols of metals.}$ $\text{Now, the stability of hydrophilic sols depends on two things- the presence of a charge and the salvation of colloidal particles.}$ $\text{On the other hand, the stability of hydrophobic sols is only because of the presence of a charge.}$ $\text{Therefore, the latter are much less stable than the former.}$ $\text{If the charge of hydrophobic sols is removed (by addition of electrolytes), then the particles present in them come closer and form aggregates, leading to precipitation.}$

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
}