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Current Question (ID: 8541)

Question:

\text{The } K_{sp} \text{ of Ag}_2\text{CrO}_4 \text{ and AgBr is } 1.1 \times 10^{-12} \text{ and } 5.0 \times 10^{-13} \text{ respectively. The molarity ratio of saturated solutions of Ag}_2\text{CrO}_4 \text{ and AgBr will be:}

Options:

1. $91.9$ (Correct)
2. $108.6$
3. $56.9$
4. $76.9$

Solution:

\text{Hint: Use the general expression of } K_{sp} \text{ to find out solubilities.} \text{Explanation:} \text{Step 1:} \text{Let } s \text{ be the solubility of } \text{Ag}_2\text{CrO}_4\text{.} \text{Ag}_2\text{CrO}_4 \rightleftharpoons 2\text{Ag}^+ + \text{CrO}_4^{2-} K_{sp} = (2s)^2 \times s \Rightarrow 1.1 \times 10^{-12} = 4s^3 \Rightarrow s = 6.5 \times 10^{-5} \text{ M} \text{Step 2:} \text{Let } s' \text{ be the solubility of } \text{AgBr}\text{.} \text{AgBr} \rightleftharpoons \text{Ag}^+ + \text{Br}^- K_{sp}' = s' \times s' \Rightarrow 5.0 \times 10^{-13} \Rightarrow s' = 7.07 \times 10^{-7} \text{ M} \text{Step 3:} \text{Find the ratio} \text{So, the ratio of molarities} = \frac{s}{s'} = \frac{6.5 \times 10^{-5}}{7.07 \times 10^{-7}} = 91.9

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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
}