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Current Question (ID: 21121)

Question:

$\text{In a saturated solution of the sparingly soluble electrolyte AgIO}_3$ $\text{(molecular mass = 283) the equilibrium is represented by:}$ $\text{AgIO}_3\text{(s)} \rightleftharpoons \text{Ag}^+\text{(aq)} + \text{IO}_3^-\text{(aq)}$ $\text{If the solubility product constant } K_{sp} \text{ of AgIO}_3 \text{ at a given temperature}$ $\text{is } 1.0 \times 10^{-8}, \text{ Calculate the mass of AgIO}_3 \text{ contained in 100 ml of its}$ $\text{saturated solution.}$

Options:

1. $28.3 \times 10^{-2} \text{ g}$
2. $2.83 \times 10^{-3} \text{ g}$
3. $1.0 \times 10^{-7} \text{ g}$
4. $4.1 \times 10^{-4} \text{ g}$

Solution:

$\text{Hint: First find solubility for 100 mL solution then multiply by molar}$ $\text{mass of AgIO}_3$ $\text{Step 1:}$ $\text{The reaction is as follows:}$ $\text{AgIO}_3 \rightarrow \text{Ag}^+ + \text{IO}_3^-$ $K_{sp} = [\text{Ag}^+][\text{IO}_3^-]$ $\text{Let the solubility be } S.$ $K_{sp} = S^2$ $\text{Calculate the concentration or solubility of AgIO}_3 \text{ as follows:}$ $1.0 \times 10^{-8} = S^2$ $\sqrt{1.0 \times 10^{-8}} = S$ $S = 1.0 \times 10^{-4} \text{ mol / L.}$ $\text{Step 2:}$ $\text{Calculate the concentration for 100 mL solution as follows:}$ $\text{For 1000 mL solution } 1.0 \times 10^{-4} \text{ mole of AgIO}_3 \text{ is present.}$ $\text{For 100 mL solution: } \frac{1 \times 10^{-4}}{1000} \times 100 \text{ mol of AgIO}_3 \text{ is present.}$ $= 1 \times 10^{-5} \text{ mol}$ $\text{Calculate the amount in g by multiply the number of mole by molar}$ $\text{mass.}$ $= 1 \times 10^{-5} \text{ mol} \times 283 \text{ g mol}^{-1}$ $= 2.83 \times 10^{-3} \text{ g}$ $\text{Hence, option 2 is the correct answer.}$

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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
}