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Current Question (ID: 18098)

Question:
$\text{The molality of a solution containing a certain solute, if there is a freezing point depression of } 0.184 \, ^\circ\text{C, is-}$ $\left(K_f = 18.4\right)$
Options:
  • 1. $0.01 \, \text{m}$
  • 2. $10 \, \text{m}$
  • 3. $0.05 \, \text{m}$
  • 4. $100 \, \text{m}$
Solution:
$\text{HINT: } \Delta T_f = K_f \times m$ $\text{Explanation:}$ $\text{The formula of depression in freezing point is as follows:}$ $\Delta T_f = K_f \times m$ $\text{Calculate the value of } m \text{ as follows:}$ $\text{or } m = \frac{\Delta T_f}{K_f} = \frac{0.184}{18.4} = 0.01 \, \text{m}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}