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Current Question (ID: 18089)

Question:

$\text{The unit of Ebullioscopic constant is:}$

Options:

1. $\text{K kg mol}^{-1} \text{ or K (molality)}^{-1}$
2. $\text{mol kg K}^{-1} \text{ or K}^{-1} \text{ (molality)}$
3. $\text{kg mol}^{-1} \text{ K}^{-1} \text{ or K}^{-1} \text{ (molality)}^{-1}$
4. $\text{K mol kg}^{-1} \text{ or K (molality)}$

Solution:

$\text{Hint: Ebullioscopic constant is same as molal elevation constant}$ $\text{As we know from elevation in boiling point that}$ $\Delta T_b = K_b m$ $K_b = \frac{\Delta T_b}{m}$ $\text{Unit of } K_b = \frac{\text{unit of } \Delta T_b}{\text{unit of } m} = \frac{\text{K}}{\text{molality}}$ $= \frac{\text{K}}{\text{mol}} \text{ kg}^{-1} = \text{K mol}^{-1} \text{ kg}$

Import JSON File

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
}