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Current Question (ID: 18204)

Question:

$\text{The conductivity of } 0.00241 \text{ M acetic acid is } 7.896 \times 10^{-5} \text{ S cm}^{-1}. \text{ If } \Lambda_m^0 \text{ for acetic acid is } 390.5 \text{ S cm}^2 \text{ mol}^{-1}, \text{ the dissociation constant will be}$

Options:

1. $2.45 \times 10^{-5} \text{ mol L}^{-1}$
2. $1.86 \times 10^{-5} \text{ mol L}^{-1}$
3. $3.72 \times 10^{-5} \text{ mol L}^{-1}$
4. $2.12 \times 10^{-5} \text{ mol L}^{-1}$

Solution:

$\text{HINT: Degree of dissociation, } \alpha = \frac{\Lambda_m}{\Lambda_m^0}$ $\text{STEP 1:}$ $\text{Conductivity (} \kappa \text{) } = 7.896 \times 10^{-5} \text{ S cm}^{-1}$ $\text{concentration } = 0.00241 \text{ mol L}^{-1}$ $\Lambda_m = \frac{\kappa}{\text{concentration}} \times 1000 = \frac{7.896 \times 10^{-5}}{0.00241} \times 1000 = 32.76 \text{ S cm}^2 \text{ mol}^{-1}$ $\Lambda_m^0 = 390.5 \text{ S cm}^2 \text{ mol}^{-1}$ $\text{STEP 2:}$ $\text{Degree of dissociation, } \alpha = \frac{\Lambda_m}{\Lambda_m^0} = \frac{32.76}{390.5} = 0.084$ $\text{Now, Dissociation constant (} K_a \text{) } = \frac{c \alpha^2}{1 - \alpha}$ $= \frac{(0.00241) \times (0.084)^2}{1 - 0.084} = 1.86 \times 10^{-5} \text{ mol L}^{-1}$

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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
}