Import Question JSON

Current Question (ID: 18086)

Question:
$1.00 \text{ g of non-electrolyte solute (molar mass } 250 \text{ g mol}^{-1}) \text{ was dissolved in } 51.2 \text{ g of benzene. If the freezing point depression constant, } K_f \text{ of benzene is } 5.12 \text{ mol}^{-1} \text{ kg K, the freezing point of benzene will be lowered by:}$
Options:
  • 1. $0.4 \text{ K}$
  • 2. $0.3 \text{ K}$
  • 3. $0.5 \text{ K}$
  • 4. $0.2 \text{ K}$
Solution:
$\Delta T_f = K_f \times \frac{W_1 \times 1000}{W_2 \times M_1}$ $\text{Step 1:}$ $W_1 = \text{Weight of the solute}$ $W_2 = \text{weight of solvent}$ $M_1 = \text{Molar mass of solute}$ $K_f = \text{Freezing point depression constant}$ $\text{Step 2:}$ $\text{Now, } \Delta T_f = 5.12 \times \frac{1 \times 1000}{51.2 \times 250}$ $\Delta T_f = 0.4 \text{ K}$

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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
}