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Current Question (ID: 18141)

Question:
$\text{A 0.0020 m aqueous solution of an ionic compound Co(NH}_3\text{)}_5\text{(NO}_2\text{)Cl freezes at } -0.0073 \, ^\circ\text{C. The number of moles of ions that 1 mol of ionic compound produces on being dissolved in water will be:}$ $\text{(}k_f = -1.86 \, ^\circ\text{C/m)}$
Options:
  • 1. $2$
  • 2. $3$
  • 3. $4$
  • 4. $1$
Solution:
$\text{Hint: } \Delta T_f = i \times k_f \times m$ $\text{Step 1:}$ $\text{It is given that,}$ $m = 0.0020 \, m$ $\Delta T_f = 0^\circ\text{C} - (-0.00732^\circ\text{C}) = 0.00732^\circ\text{C}$ $\Delta T_f = i \times k_f \times m$ $\Delta T_f = \text{Depression in freezing point, } m = \text{molality and } k_f = 1.86^\circ\text{C/m}$ $\text{Step 2:}$ $\text{Calculate the value of } i \text{ as follows:}$ $i = \frac{\Delta T_f}{k_f \times m}$ $i = \frac{0.00732}{1.86 \times 0.0020}$ $= 1.96 \approx 2$ $\text{As } i > 1, \text{ it means dissociation of the solute occurs i.e. from one mole of ionic compound, 2 moles of ions are produced.}$

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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
}