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Current Question (ID: 18707)

Question:

$\text{A solution containing 2.675 g of CoCl}_3\cdot 6\text{NH}_3 \text{ (molar mass = 267.5 g mol}^{-1}\text{) is passed through a cation exchanger. The chloride ions obtained in the solution were treated with an excess of AgNO}_3 \text{ to give 4.78 g of AgCl. Formula of the complex is (Atomic mass of Ag = 108u):}$

Options:

1. $[\text{CoCl(NH}_3)_5]\text{Cl}_2$
2. $\left[ \text{Co} \left( \text{NH}_3 \right)_6 \right] \text{Cl}_3$
3. $[\text{CoCl}_2(\text{NH}_3)_4]\text{Cl}$
4. $[\text{CoCl}_3(\text{NH}_3)_3]$

Solution:

$\text{HINT: One mole of Cl}^- \text{ reacts with one mole of AgNO}_3 \text{ to give one mole of AgCl.}$ $\text{Explanation: Only the Cl}^- \text{ present outside the coordination sphere are ionisable. Such Cl}^- \text{ can be precipitated out.}$ $\text{Moles of CoCl}_3\cdot 6\text{NH}_3 = \frac{W}{\text{M.W}} = \frac{2.675}{267.5} = 0.01 \text{ mol}$ $\text{AgNO}_3 + \text{Cl}^- \rightarrow \text{AgCl} \downarrow + \text{NO}_3^- \text{ (aq)}$ $\text{Moles of AgCl} = \frac{W}{\text{M.W}} = \frac{4.78}{143.5} = 0.03 \text{ mol}$ $0.01 \text{ mol CoCl}_3\cdot 6\text{NH}_3 \text{ gives } = 0.03 \text{ mol AgCl}$ $= \frac{0.03}{0.01} = 3 \text{ mol}$ $\text{Thus, the formula of the complex is } \left[ \text{Co} \left( \text{NH}_3 \right)_6 \right] \text{Cl}_3$

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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
}