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Current Question (ID: 21410)

Question:

$\text{For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?}$ $1. \ [\text{Co} (\text{H}_2\text{O})_6]\text{Cl}_3$ $2. \ [\text{Co} (\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O}$ $3. \ [\text{Co} (\text{H}_2\text{O})_4\text{Cl}_2]\text{Cl} \cdot 2\text{H}_2\text{O}$ $4. \ [\text{Co} (\text{H}_2\text{O})_3\text{Cl}_3] \cdot 3\text{H}_2\text{O}$

Options:

1. $[\text{Co} (\text{H}_2\text{O})_6]\text{Cl}_3$
2. $[\text{Co} (\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O}$
3. $[\text{Co} (\text{H}_2\text{O})_4\text{Cl}_2]\text{Cl} \cdot 2\text{H}_2\text{O}$
4. $[\text{Co} (\text{H}_2\text{O})_3\text{Cl}_3] \cdot 3\text{H}_2\text{O}$

Solution:

$\text{Hint: } \Delta T_f \text{ (Freezing point depression) } \propto \ n \text{ (no. of particles)}$ $\text{As the number of particles in the solution increases, the freezing point of the solution is decreased.}$ $[\text{Co} (\text{H}_2\text{O})_6]\text{Cl}_3 \text{ will generate 4 ions.}$ $[\text{Co} (\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O} \text{ will generate three ions}$ $[\text{Co} (\text{H}_2\text{O})_4\text{Cl}_2]\text{Cl} \cdot 2\text{H}_2\text{O} \text{ will generate two ions}$ $[\text{Co} (\text{H}_2\text{O})_3\text{Cl}_3] \cdot 3\text{H}_2\text{O} \text{ will generate one ion.}$ $\text{As the number of ions increases, the freezing point of the solution decreases.}$ $\text{Hence, } [\text{Co} (\text{H}_2\text{O})_3\text{Cl}_3] \cdot 3\text{H}_2\text{O} \text{ solution has the least number of ions, so it will have the highest freezing point.}$

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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
}