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Current Question (ID: 21373)

Question:

$\text{The limiting molar conductivities of NaI, NaNO}_3, \text{ and AgNO}_3 \text{ are } 12.7 \ S \ m^2 \ mole^{-2}, 12 \ S \ m^2 \ mole^{-2} \text{ and } 13.3 \ S \ m^2 \ mole^{-2} \text{ respectively (all at } 25^\circ\text{C). The limiting molar conductivity of AgI at this temperature is:}$

Options:

1. $14 \ S \ m^2 \ mole^{-2}$
2. $16 \ S \ m^2 \ mole^{-2}$
3. $48 \ S \ m^2 \ mole^{-2}$
4. $12.6 \ S \ m^2 \ mole^{-2}$

Solution:

$\text{Given } \lambda_m^\infty(\text{NaI}) = 12.7 \ S \ m^2 \ mole^{-2}$ $\lambda_m^\infty(\text{AgNO}_3) = 13.3 \ S \ m^2 \ mole^{-2}$ $\lambda_m^\infty(\text{NaNO}_3) = 12 \ S \ m^2 \ mole^{-2}$ $\lambda_m^\infty(\text{AgI}) = \lambda_m^\infty(\text{AgNO}_3) + \lambda_m^\infty(\text{NaI}) - \lambda_m^\infty(\text{NaNO}_3)$ $= 13.3 + 12.7 - 12$ $= 14 \ S \ m^2 \ mole^{-2}$

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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
}