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Current Question (ID: 18434)

Question:

$\text{Which mixture of the solutions will lead to the formation of a negatively charged colloidal [AgI]I}^{-} \text{ sol?}$

Options:

1. $50 \text{ mL of } 0.1 \text{ M AgNO}_3 + 50 \text{ mL of } 0.1 \text{ M KI}$
2. $50 \text{ mL of } 1 \text{ M AgNO}_3 + 50 \text{ mL of } 0.5 \text{ M KI}$
3. $50 \text{ mL of } 1 \text{ M AgNO}_3 + 50 \text{ mL of } 2 \text{ M KI}$
4. $50 \text{ mL of } 2 \text{ M AgNO}_3 + 50 \text{ mL of } 1.5 \text{ M KI}$

Solution:

$\text{HINT: KI should be more than AgNO}_3$ $\text{The electrical charge on the colloidal particle is explained by the process of preferential adsorption of ions from the solution.}$ $\text{During the preparation of colloidal sol, an ionic colloidal adsorb ions common to its lattice.}$ $\text{Colloidal sol AgI is prepared by adding KI solution to the AgNO}_3 \text{ solution till KI is in slight excess, iodide ion (I}^{-}) \text{ will get adsorbed on the surface of AgI particles thereby giving them negative charge.}$ $\text{In (3) option millimoles of KI is more than AgNO}_3, \text{ so this will result in the formation of negatively charged [AgI]I}^{-} \text{ sol.}$ $\text{As per NTA both 2, 3 can be the right ans..}$

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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
}