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Current Question (ID: 8535)

Question:

$\text{The molar solubility of CaF}_2 \text{ (K}_{\text{sp}} = 5.3 \times 10^{-11}) \text{ in } 0.1 \text{ M solution of NaF will be:}$

Options:

1. $5.3 \times 10^{-11} \text{ mol L}^{-1}$
2. $5.3 \times 10^{-8} \text{ mol L}^{-1}$
3. $5.3 \times 10^{-9} \text{ mol L}^{-1}$
4. $5.3 \times 10^{-10} \text{ mol L}^{-1}$

Solution:

\text{Hint: } K_{\text{sp}} = [\text{Ca}^{2+}][\text{F}^{-}]^2 \text{Step 1:} \text{The reaction is as follows:} \text{CaF}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{F}^{-} \quad \text{X} \quad \quad \quad 2\text{X} \text{The dissociation of NaF is as shown below:} \text{NaF} \rightarrow 0.1\text{Na}^{+} + 0.1\text{F}^{-} \text{Thus total fluoride ion concentration is } 2x + 0.1 \text{ M but } 2x \ll 0.1\text{.} \text{Hence, } 2x \text{ is neglected and the total fluoride ion concentration is } 0.1 \text{ M.} \text{Step 2:} \text{The expression for the solubility product is as follows:} K_{\text{sp}} = [\text{Ca}^{2+}][\text{F}^{-}]^2 5.3 \times 10^{-11} = x(0.1)^2 x = 5.3 \times 10^{-9} \text{ M}

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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
}