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Current Question (ID: 8330)

Question:
$\text{The entropy change in the fusion of one mole of a solid melting at } 27^{\circ} \text{C is:}$ $(\text{the latent heat of fusion is } 2930 \text{ J mol}^{-1})$
Options:
  • 1. $9.77 \text{ JK}^{-1}\text{mol}^{-1}$
  • 2. $19.73 \text{ JK}^{-1}\text{mol}^{-1}$
  • 3. $2930 \text{ JK}^{-1}\text{mol}^{-1}$
  • 4. $108.5 \text{ JK}^{-1}\text{mol}^{-1}$
Solution:
$\text{Hint: } \Delta\text{S} = \frac{\Delta\text{H}_f}{\text{T}}$ $\text{Step 1: Identify the known values}$ $\text{Given:}$ $\text{- Latent heat of fusion (}\Delta\text{H}_f) = 2930 \text{ J mol}^{-1}$ $\text{- Temperature (T) = } 27^{\circ}\text{C}$ $\text{Step 2: Convert the temperature to Kelvin}$ $\text{Temperature in Kelvin = Temperature in } ^{\circ}\text{C } + 273.15$ $\text{T = } 27 + 273.15 = 300.15 \text{ K}$ $\text{For simplicity, we can approximate this to 300 K, which appears to be what was done in the solution image.}$ $\text{Step 3: Calculate the entropy change using the formula}$ $\text{The entropy change during fusion (}\Delta\text{S}_f\text{) can be calculated using:}$ $\Delta\text{S}_f = \frac{\Delta\text{H}_f}{\text{T}}$ $\text{Where:}$ $\Delta\text{S}_f = \text{entropy change of fusion}$ $\Delta\text{H}_f = \text{enthalpy change of fusion (latent heat of fusion)}$ $\text{T = temperature in Kelvin}$ $\text{Substituting the values:}$ $\Delta\text{S}_f = \frac{2930 \text{ J mol}^{-1}}{300 \text{ K}} = 9.77 \text{ J mol}^{-1} \text{ K}^{-1}$ $\text{Therefore, the entropy change in the fusion of one mole of the solid is } 9.77 \text{ J K}^{-1} \text{mol}^{-1}$ $\text{This makes physical sense because fusion (melting) is a process where the solid changes to a liquid, resulting in increased disorder and hence a positive entropy change.}$

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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
}