Import Question JSON

Current Question (ID: 21438)

Question:
$\text{If the freezing point of } 0.5 \text{ w/w } \% \text{ KCl (aq) solution is } -0.24^\circ\text{C.}$ $\text{The percentage of dissociation of KCl in the solution is:}$ $[K_f=1.86 \text{ K-kg/mol}]$
Options:
  • 1. $99 \%$
  • 2. $67 \%$
  • 3. $79 \%$
  • 4. $91 \%$
Solution:
$\text{Hint: } \Delta T_f = iK_fm$ $\Delta T_f = iK_fm$ $0.24 = i \times 1.86 \left[ \frac{0.5 \times 1000}{74.5 \times 99.5} \right]$ $i = 1.91$ $i = (1 + (n - 1)\alpha); n = 2 \text{ for } \text{KCl}$ $\text{so, } 1.91 = 1 + \alpha; \alpha = 0.91 = 91\%$

Import JSON File

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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
}