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Current Question (ID: 21085)

Question:

$\text{If solubility product of } \text{Zr}_3(\text{PO}_4)_4 \text{ is denoted by } K_{sp} \text{ and its molar solubility is denoted by } S, \text{ then which of the following relation between } S \text{ and } K_{sp} \text{ is correct?}$

Options:

1. $S = \left( \frac{K_{sp}}{216} \right)^{1/7}$
2. $S = \left( \frac{K_{sp}}{6912} \right)^{1/7}$
3. $S = \left( \frac{K_{sp}}{144} \right)^{1/6}$
4. $S = \left( \frac{K_{sp}}{929} \right)^{1/9}$

Solution:

$\text{Hint: Use the formula, that is, } K_{sp} = (3S)^3 (4S)^4$ $\text{The reaction of } \text{Zr}_3(\text{PO}_4)_4 \text{ is as follows:}$ $\text{Zr}_3(\text{PO}_4)_4 \rightleftharpoons 3\text{Zr}^{4+} + 4\text{PO}_4^{3-}$ $\text{The } K_{sp} \text{ of } \text{Zr}_3(\text{PO}_4)_4 \text{ is as follows:}$ $K_{sp} = (3S)^3 (4S)^4$ $K_{sp} = 6912S^7$ $S = \left( \frac{K_{sp}}{6912} \right)^{1/7}$

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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
}