Import Question JSON

Current Question (ID: 21144)

Question:

$\text{The solubility product of } \text{PbI}_2 \text{ is } 8.0 \times 10^{-9}. \text{ The solubility of lead iodide in } 0.1 \text{ molar solution of lead nitrate is } x \times 10^{-6} \text{ mol/L. The value of } x \text{ is:}$ $\text{(Rounded off to the nearest integer) } [\text{Given: } \sqrt{2} = 1.41]$

Options:

1. 154
2. 423
3. 282
4. 141

Solution:

$\text{Hint: The } K_{sp} \text{ formula for } \text{PbI}_2 \text{ is } [\text{Pb}^{+2}][\text{I}^-]^2$ $\text{Step 1:}$ $\text{Given: } [K_{sp}]_{\text{PbI}_2} = 8 \times 10^{-9}$ $\text{To calculate: Solubility of } \text{PbI}_2 \text{ in } 0.1 \text{ M sol of } \text{Pb(NO}_3)_2$ $\text{Pb(NO}_3)_2 \rightarrow \text{Pb}^{+2}_{(aq)} + 2\text{NO}_3^-_{(aq)}$ $0.1\text{M} \quad - \quad 0.1\text{M} \quad - \quad 0.2\text{M}$ $\text{PbI}_2_{(s)} \rightleftharpoons \text{Pb}^{+2}_{(aq)} + 2\text{I}^-_{(aq)}$ $\approx \frac{s}{0.1} + \frac{s}{0.1}$ $\text{Step 2:}$ $\text{Now, } K_{sp} = 8 \times 10^{-9} = [\text{Pb}^{+2}][\text{I}^-]^2$ $\Rightarrow 8 \times 10^{-9} = 0.1 \times (2s)^2$ $\Rightarrow 8 \times 10^{-8} = 4s^2 \left( s = \sqrt{2} \times 10^{-4} \right)$ $\Rightarrow s = 141 \times 10^{-6}\text{M}$ $\Rightarrow x = 141$

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
}