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Current Question (ID: 18207)

Question:

$\text{The specific conductance of a 0.01 M solution of a weak monobasic acid is } 0.20 \times 10^{-3} \text{ S cm}^{-1}. \text{ The dissociation constant of the acid is-}$ $\text{[Given } \Lambda_{\text{HA}}^{\infty} = 400 \text{ S cm}^2 \text{ mol}^{-1}]$

Options:

1. $5 \times 10^{-2}$
2. $2.5 \times 10^{-5}$
3. $5 \times 10^{-4}$
4. $2.2 \times 10^{-11}$

Solution:

$\text{HINT: } \alpha = \frac{\Lambda_M}{\Lambda_M^{\infty}}$ $\text{Step 1:}$ $\text{Calculate the molar conductivity as follows:}$ $\Lambda_M = \frac{k \times 1000}{M}$ $= \frac{0.20 \times 10^{-3} \times 1000}{0.01}$ $\Lambda_M = 20$ $\text{Step 2:}$ $\text{The relation between dissociation coefficient with molar conductance and limiting molar conductance is as follows:}$ $\alpha = \frac{\Lambda_m}{\Lambda_m^{\infty}}$ $\alpha = \frac{20}{400} = \frac{1}{20}$ $\text{Step 3:}$ $\text{Calculate the value of the dissociation constant as follows:}$ $K = C \alpha^2 = 0.01 \times \frac{1}{400} = 2.5 \times 10^{-5}$

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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
}