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Current Question (ID: 18112)

Question:

$\text{Compound PdCl}_4\cdot 6\text{H}_2\text{O is a hydrated complex, 1 molal aqueous solution of it has freezing point 269.28 K. Assuming 100 \% ionization of complex, the molecular formula of the complex is-}$ $\left(K_f \text{ for water} = 1.86 \text{ K kg mole}^{-1}\right)$

Options:

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

Solution:

$\text{Hint: } i \text{ can be calculated using } \Delta T_f$ $\text{Step 1:}$ $\text{Given: Molality} = 1 \text{ m,}$ $\Delta T_f = T_f(\text{pure}) - T_f(\text{solution})$ $= 273.00 - 269.28$ $= 3.72$ $K_f = 1.86 \text{ K kg mole}^{-1}$ $\text{Step 2:}$ $\text{We know that}$ $\Delta T_f = i \times K_f \times m$ $3.72 = i \times 1.86 \times 1$ $i = 2$ $\text{Hence, complex must give 2 ions. So, complex is [Pd(H}_2\text{O)}_3\text{Cl}_3]\cdot \text{Cl}\cdot 3\text{H}_2\text{O}$

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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
}