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Current Question (ID: 18441)

Question:
$1 \text{ mol of } [\text{AgI}]^- \text{ can be coagulated by -}$ $1. \text{ 1 mol of } \text{Pb(NO}_3)_2$ $2. \frac{1}{2} \text{ mol of } \text{Pb(NO}_3)_2$ $3. \frac{2}{3} \text{ mol of } \text{Pb(NO}_3)_2$ $4. \text{ None of the above.}$
Options:
  • 1. $1 \text{ mol of } \text{Pb(NO}_3)_2$
  • 2. $\frac{1}{2} \text{ mol of } \text{Pb(NO}_3)_2$
  • 3. $\frac{2}{3} \text{ mol of } \text{Pb(NO}_3)_2$
  • 4. $\text{None of the above.}$
Solution:
$[\text{AgI}]^- \text{ is a negatively charged sol, so it will be coagulated by } \text{Pb}^{2+}. \text{ Write down the balanced reaction to calculate number of moles.}$ $\text{Pb(NO}_3)_2 + 2[\text{AgI}]^- \rightarrow \text{PbI}_2 \downarrow + \text{AgI} \downarrow$ $\text{Thus 2 moles of } [\text{AgI}]^- \text{ are coagulated by 1 mole of } \text{Pb}^{2+} \text{ i.e., 1 mole of } \text{Pb(NO}_3)_2$ $\text{Hence,}$ $2[\text{AgI}]^- \equiv \text{Pb(NO}_3)_2 \therefore [\text{AgI}]^- \equiv \frac{1}{2} \text{Pb(NO}_3)_2$

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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
}