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Current Question (ID: 8533)

Question:

$\text{If the solubility of a } \text{M}_2\text{S salt is } 3.5 \times 10^{-6} \text{ mol litre}^{-1}, \text{ then the solubility product of } \text{M}_2\text{S will be:}$

Options:

1. $1.7 \times 10^{-6} \text{ mol}^3 \text{ litre}^{-3}$
2. $1.7 \times 10^{-16} \text{ mol}^3 \text{ litre}^{-3}$
3. $1.7 \times 10^{-18} \text{ mol}^3 \text{ litre}^{-3}$
4. $1.7 \times 10^{-12} \text{ mol}^3 \text{ litre}^{-3}$

Solution:

\text{Hint: } K_{\text{sp}} = [\text{M}^{+}]^2 [\text{S}^{2-}] \text{Step 1:} \text{The dissociation reaction of } \text{M}_2\text{S} \text{ is as follows:} \text{M}_2\text{S} \rightleftharpoons 2\text{M}^{+} + \text{S}^{2-} \text{The } K_{\text{sp}} \text{ formula for } \text{M}_2\text{S} \text{ is as follows:} K_{\text{sp}} = [\text{M}^{+}]^2 [\text{S}^{2-}] K_{\text{sp}} = (2\text{S})^2 (\text{S}) K_{\text{sp}} = 4\text{S}^3 \text{Step 2:} \text{Calculate the value of solubility product is as follows:} K_{\text{sp}} = 4\text{S}^3 K_{\text{sp}} = 4 \times (3.5 \times 10^{-6})^3 K_{\text{sp}} = 1.7 \times 10^{-16}

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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
}