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Current Question (ID: 10793)

Question:

$\text{The value of the coefficient of volume expansion of glycerin is } 5 \times 10^{-4} \text{ K}^{-1}\text{. The fractional change in the density of glycerin for a temperature increase of } 40°\text{C will be:}$

Options:

1. $0.015$
2. $0.020$ (Correct)
3. $0.025$
4. $0.010$

Solution:

\text{Given: Coefficient of volume expansion } \gamma = 5 \times 10^{-4} \text{ K}^{-1} \text{Temperature increase } \Delta T = 40°\text{C} \text{For volume expansion: } \rho_T = \rho_0(1 - \gamma \Delta T) \text{The fractional change in density is:} \frac{\rho_T}{\rho_0} = (1 - \gamma \Delta T) \text{Rearranging: } 1 - \frac{\rho_T}{\rho_0} = \gamma \Delta T \frac{\rho_0 - \rho_T}{\rho_0} = \gamma \Delta T = 5 \times 10^{-4} \times 40 = 0.020 \text{Therefore, the fractional decrease in density is } 0.020 \text{ or } 2.0\%

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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
}