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Current Question (ID: 18460)

Question:

$\text{Match the items given in Column I and Column II.}$ $\text{Column I}$ \quad \text{Column II}$ $\text{A. Protective colloid} \quad 1. \text{FeCl}_3 + \text{NaOH}}$ $\text{B. Liquid-liquid colloid} \quad 2. \text{Lyophilic colloids}}$ $\text{C. Positively charged colloid} \quad 3. \text{Emulsion}}$ $\text{D. Negatively charged colloid} \quad 4. \text{FeCl}_3 + \text{hot water}}$ $\text{Codes}$

Options:

1. $1. \quad 3 \quad 2 \quad 4 \quad 1$
2. $2. \quad 1 \quad 4 \quad 3 \quad 2$
3. $3. \quad 2 \quad 3 \quad 4 \quad 1$
4. $4. \quad 4 \quad 1 \quad 3 \quad 2$

Solution:

$\text{Hint: When ferric chloride is added to NaOH solution a negatively charged sol is obtained with adsorption of OH}^- \text{ ions.}$ $\text{The protective colloid is a type of lyophilic (water-loving) colloid that is used to protect the lyophobic colloids from precipitating in an electrolytic solution.}$ $\text{An emulsion is a mixture of two or more liquids that are normally immiscible (unmixable or unblendable) owing to liquid-liquid phase separation.}$ $\text{If FeCl}_3 \text{ is added to the excess of hot water, a positively charged sol of hydrated ferric oxide is formed due to the adsorption of Fe}^{3+} \text{ ions.}$ $\text{When ferric chloride is added to NaOH solution a negatively charged sol is obtained with adsorption of OH}^- \text{ ions.}$ $\text{hence, option third is the correct answer.}$

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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
}