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Current Question (ID: 18201)

Question:

$\text{At 25 °C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is } 9.54 \, \Omega^{-1} \text{ cm}^2 \text{ mol}^{-1} \text{ and at infinite dilution its molar conductance is } 238 \, \Omega^{-1} \text{ cm}^2 \text{ mol}^{-1}. \text{ The degree of ionisation of ammonium hydroxide at the same concentration and temperature is:}$

Options:

1. $2.080 \%$
2. $20.800 \%$
3. $4.008 \%$
4. $40.800 \%$

Solution:

$\text{HINT: } \alpha = \frac{\Lambda_c}{\Lambda_c^\infty}$ $\text{Explanation:}$ $\text{STEP 1: Given, molar conductance at 0.1 M concentration,}$ $\Lambda_c = 9.54 \, \Omega^{-1} \text{ cm}^2 \text{ mol}^{-1}$ $\text{Molar conductance at infinite dilution,}$ $\Lambda_c^\infty = 238 \, \Omega^{-1} \text{ cm}^2 \text{ mol}^{-1}.$ $\text{STEP 2: We know that,}$ $\text{Degree of ionisation, } \alpha = \frac{\Lambda_c}{\Lambda_c^\infty}$ $= \frac{9.54}{238} \times 100 = 4.008\%$

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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
}