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Current Question (ID: 18671)

Question:

$\text{If } \frac{1}{3} \text{ of the total chlorine of the compound is precipitated by adding } \text{AgNO}_3 \text{ to its aqueous solution, then the most likely structure of } \text{CrCl}_3 \cdot 6\text{H}_2\text{O} \text{ among the following is:}$

Options:

1. $\text{CrCl}_3 \cdot 6\text{H}_2\text{O}$
2. $[\text{Cr(H}_2\text{O})_3\text{Cl}_3](\text{H}_2\text{O})_3$
3. $[\text{CrCl}_2(\text{H}_2\text{O})_4]\text{Cl} \cdot 2\text{H}_2\text{O}$
4. $[\text{CrCl.(H}_2\text{O})_5]\text{Cl}_2 \cdot \text{H}_2\text{O}$

Solution:

$\text{Step 1: What is the reaction of } \text{AgNO}_3 \text{ with } \text{Cl}^-? \newline \text{AgNO}_3 + \text{Cl}^- \rightarrow \text{AgCl} \downarrow_{\text{White}}$ $\text{Step 2: How many chlorine will come out with } \text{AgNO}_3? \newline 1. \text{ Cannot be ascertained} \newline 2. \text{ 0 Cl}^- \text{ will be generated} \newline 3. \text{ 1 Cl}^- \text{ will be generated} \newline 4. \text{ 2 Cl}^- \text{ will be generated}$ $\text{Step 3: Which are those chlorine which come out with } \text{AgNO}_3? \newline \text{Chlorine present outside the coordination sphere are going to get ionized.}$ $[\text{CrCl}_2(\text{H}_2\text{O})_4]\text{Cl} \cdot 2\text{H}_2\text{O} \xrightarrow{\text{AgNO}_3} \text{AgCl (ppt)}$ $\text{Out of the coordination sphere, only one chloride ion is present. Ions that is present out of coordination sphere only participate in precipitation.}$

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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
}