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Current Question (ID: 18203)

Question:

$\lambda_m^0(\text{H}_2\text{SO}_4) = x \ S \ \text{cm}^2 \ \text{mol}^{-1}$ $\lambda_m^0(\text{K}_2\text{SO}_4) = y \ S \ \text{cm}^2 \ \text{mol}^{-1}$ $\lambda_m^0(\text{CH}_3\text{COOK}) = z \ S \ \text{cm}^2 \ \text{mol}^{-1}$ $\text{From the data given above, it can be concluded that } \lambda_m^0 \text{ in } (S \ \text{cm}^2 \ \text{mol}^{-1}) \text{ for CH}_3\text{COOH will be:}$

Options:

1. $x - y + 2z$
2. $x + y + z$
3. $x - y + z$
4. $\frac{(x-y)}{2} + z$

Solution:

$\text{HINT: Application of Kohlrausch's law}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COO}^- + \text{H}^+ \ldots (1)$ $\text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-} \ldots (2)$ $\text{K}_2\text{SO}_4 \rightarrow 2\text{K}^+ + \text{SO}_4^{2-} \ldots (3)$ $\text{CH}_3\text{COOK} \rightarrow \text{CH}_3\text{COO}^- + \text{K}^+ \ldots (4)$ $\text{Step 2:}$ $\text{According to Kohlrausch's law -}$ $\lambda^0_{\text{CH}_3\text{COOH}} = \lambda^0_{\text{CH}_3\text{COO}^-} + \lambda^0_{\text{H}^+}$ $\text{eq. (1) = eq. (4) + eq. } \frac{(2)}{2} - \text{eq. } \frac{(3)}{2}$ $\therefore \lambda^0_{\text{CH}_3\text{COOH}} = z + \frac{x}{2} - \frac{y}{2}$ $\lambda^0_{\text{CH}_3\text{COOH}} = \frac{(x-y)}{2} + z \ (S \times \text{cm}^2 \text{mol}^{-1})$

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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
}