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Current Question (ID: 18209)

Question:

$\text{Molar conductivities (}\Lambda_m^\circ\text{) at infinite dilution of NaCl, HCl, and CH}_3\text{COONa are 126.4, 425.9, and 91.0 S cm}^2\text{ mol}^{-1}\text{ respectively.}$ $\Lambda_m^\circ\text{ for CH}_3\text{COOH will be:}$

Options:

1. $180.5 \text{ S cm}^2\text{ mol}^{-1}$
2. $290.8 \text{ S cm}^2$
3. $390.5 \text{ S cm}^2\text{ mol}^{-1}$
4. $425.5 \text{ S cm}^2\text{ mol}^{-1}$

Solution:

$\text{Use application of Kohlrausch's law.}$ $\text{Step 1:}$ $\text{It is given:}$ $\Lambda_m^\circ(\text{NaCl}) = 126.4 \text{ S cm}^2\text{ mol}^{-1}, \Lambda_m^\circ(\text{HCl}) = 425.9 \text{ S cm}^2\text{ mol}^{-1}, \Lambda_m^\circ(\text{CH}_3\text{COONa}) = 91.0 \text{ S cm}^2\text{ mol}^{-1}$ $\text{We need to calculate } \Lambda_m^\circ(\text{CH}_3\text{COOH})$ $\text{CH}_3\text{COONa} + \text{HCl} \rightarrow \text{CH}_3\text{COOH} + \text{NaCl}$ $\text{Step 2:}$ $\Lambda_m^\circ(\text{CH}_3\text{COOH}) = \Lambda_m^\circ(\text{CH}_3\text{COONa}) + \Lambda_m^\circ(\text{HCl}) - \Lambda_m^\circ(\text{NaCl})$ $= 91.0 + 425.9 - 126.4 = 390.5 \text{ S cm}^2\text{ mol}^{-1}$

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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
}