Import Question JSON

Current Question (ID: 8342)

Question:

\text{The molar entropy of the vapourization of acetic acid is } 14.4 \text{ cal K}^{-1} \text{ mol}^{-1} \text{ at its boiling point } 118°\text{C}. \text{The latent heat of vapourization of acetic acid is:} \text{1. } 49 \text{ cal g}^{-1} \quad \text{2. } 64 \text{ cal g}^{-1} \text{3. } 94 \text{ cal g}^{-1} \quad \text{4. } 84 \text{ cal g}^{-1}

Options:

1. $49 \text{ cal g}^{-1}$
2. $64 \text{ cal g}^{-1}$
3. $94 \text{ cal g}^{-1}$
4. $84 \text{ cal g}^{-1}$

Solution:

$\text{Hint:}$ $\Delta\text{s} = \frac{\Delta\text{H}}{\text{T (K)}}$ $\text{To calculate the latent heat of vaporization, we need to use the relationship between entropy change and enthalpy change during a phase transition at constant temperature:}$ $\Delta\text{s} = \frac{\Delta\text{H}}{\text{T}}$ $\text{Where:}$ $\Delta\text{s = entropy of vaporization = 14.4 cal K}^{-1} \text{ mol}^{-1}$ $\Delta\text{H = latent heat of vaporization (in cal mol}^{-1}\text{)}$ $\text{T = boiling point temperature in Kelvin = 118}^{\circ}\text{C + 273 = 391 K}$ $\text{Rearranging to solve for}$ $\Delta\text{H:}$ $\Delta\text{H} = \Delta\text{s} \times \text{T}$ $\Delta\text{H} = 14.4 \text{ cal K}^{-1} \text{ mol}^{-1} \times 391 \text{ K}$ $\Delta\text{H} = 5630 \text{ cal mol}^{-1}$ $\text{Now, we need to convert this to cal g}^{-1}\text{. The molecular weight of acetic acid (CH}_{3}\text{COOH) is 60 g mol}^{-1}\text{:}$ $\Delta\text{H in cal g}^{-1} = \frac{5630 \text{ cal mol}^{-1}}{60 \text{ g mol}^{-1}} = 94 \text{ cal g}^{-1}$ $\text{Therefore, the latent heat of vaporization of acetic acid is 94 cal g}^{-1}\text{, which corresponds to option 3.}$

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
}