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Current Question (ID: 21356)

Question:

$\text{For a given cell, a 0.1 molar solution has a resistance of } 20 \, \Omega \text{ and}$ $\text{molar conductivity of } 0.154 \times 10^{-3} \, S \, \text{cm}^2 \, \text{mol}^{-1}.$ $\text{The value of the cell constant is:}$

Options:

1. $3.08 \times 10^{-7} \, \text{cm}^{-1}$
2. $30.8 \times 10^{-7} \, \text{cm}^{-1}$
3. $0.308 \times 10^{-9} \, \text{cm}^{-1}$
4. $4.08 \times 10^{-6} \, \text{cm}^{-1}$

Solution:

$\text{Hint: } k = C \times x$ $\text{Step 1:}$ $\text{First calculate the value of conductivity (kappa) by using the molar}$ $\text{conductivity formula as follows:}$ $\lambda_m = \frac{k \times 1000}{M}$ $0.154 \times 10^{-3} = \frac{k \times 1000}{0.1}$ $k = 0.154 \times 10^{-7} \, S \, \text{cm}^{-1}$ $\text{Step 2:}$ $\text{Now calculate the value of cell constant as follows:}$ $k = \left( \frac{\ell}{a} \right) \frac{1}{R}$ $\text{Cell constant } \left( \frac{\ell}{a} \right) = k \times R$ $= 0.154 \times 10^{-7} \times 20$ $= 3.08 \times 10^{-7} \, \text{cm}^{-1}$

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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
}