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Current Question (ID: 21386)

Question:

$\text{Determine the cell constant of a conductivity cell containing a } 0.01 \text{ M KCl solution at } 298 \text{ K.}$ $\text{The given data includes a resistance of } 1750 \, \Omega$ $\text{and a conductivity of } 0.152 \times 10^{-3} \, \text{S cm}^{-1}.$

Options:

1. $266 \times 10^{-3} \, \text{m}^{-1}$
2. $166 \times 10^{-3} \, \text{cm}^{-1}$
3. $266 \times 10^{-3} \, \text{cm}^{-1}$
4. $166 \times 10^{-3} \, \text{m}^{-1}$

Solution:

$\text{Hint: } \mathcal{G} = \frac{1}{R} = \frac{A}{\rho l} = \kappa \frac{A}{l}$ $\text{Molarity of KCl solution} = 0.1 \, M$ $\text{Resistance} = 1750 \, \Omega$ $\text{Conductivity} = 0.152 \times 10^{-3} \, \text{S cm}^{-1}$ $\text{Conductivity} = \frac{\text{Cell constant}}{\text{Resistance}}$ $\therefore \text{Cell constant} = 0.152 \times 10^{-3} \times 1750 = 266 \times 10^{-3} \, \text{cm}^{-1}$

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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
}