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Current Question (ID: 18136)

Question:

$\text{The electrolyte having the same value of Van't Hoff factor (i) as that of } \text{Al}_2(\text{SO}_4)_3 \text{ (if all are 100\% ionized) is:}$

Options:

1. $\text{K}_2\text{SO}_4$
2. $\text{K}_3[\text{Fe(CN)}_6]$
3. $\text{Al(NO}_3)_3$
4. $\text{K}_4[\text{Fe(CN)}_6]$

Solution:

$\text{Hint: Van't Hoff factor (i) is the number of ions generated by compounds after dissociation or association.}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{Calculate the value of i for } \text{Al}_2(\text{SO}_4)_3 \text{ as follows:}$ $\text{For the equation,}$ $\text{Al}_2(\text{SO}_4)_3 \rightarrow 2\text{Al}^{3+} + 3\text{SO}_4^{2-}$ $\text{Value of Van't Hoff factor, i = 5}$ $\text{Step 2:}$ $\text{Calculate the value of i for other compounds given in the options as follows:}$ $(a) \text{K}_2\text{SO}_4 \rightarrow 2\text{K}^{+} + \text{SO}_4^{2-}$ $\text{Value of i=3}$ $(b) \text{K}_3[\text{Fe(CN)}_6] \rightarrow 3\text{K}^{+} + [\text{Fe(CN)}_6]^{3-}$ $\text{Value of i=4}$ $(c) \text{Al(NO}_3)_3 \rightarrow \text{Al}^{3+} + 3\text{NO}_3^{-}$ $\text{Value of i=4}$ $(d) \text{K}_4[\text{Fe(CN)}_6] \rightarrow 4\text{K}^{+} + [\text{Fe(CN)}_6]^{4-}$ $\text{Value of i=5}$ $\text{Hence, } \text{K}_4[\text{Fe(CN)}_6] \text{ has the same value of i as that of } \text{Al}_2(\text{SO}_4)_3.$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}