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Current Question (ID: 21397)

Question:

$\text{When 12.2 g of benzoic acid is dissolved in 100 g of water, the freezing point of the solution was found to be }$ $-0.93 \, ^\circ \text{C. The number } (n) \text{ of benzoic acid molecules associated (assuming 100\% association) is:}$ $\left( K_f(\text{H}_2\text{O}) = 1.86 \, \text{K kg mol}^{-1} \right)$

Options:

1. 3 (Correct)
2. 1
3. 2
4. 5

Solution:

$\text{Hint: The relation between } i, n \text{ and } \alpha \text{ for association reaction is }$ $i = 1 + \left( \frac{1}{n} - 1 \right) \alpha$ $\text{Step 1:}$ $\text{First, calculate the value of } i \text{ using the following formula}$ $\Delta T_f = i \times k_f \times m$ $T_f^0 - T_f = i \times K_f \times \frac{\text{amount of benzoic acid}}{\text{molar mass of benzoic acid}} \times \frac{1000}{\text{volume of soln}}$ $0 - (-0.93) = i \times 1.86 \times \frac{12.2}{122 \times 100} \times 1000$ $i = \frac{0.93}{1.86} = 0.5$ $\text{Step 2:}$ $\text{Calculate the value of } n \text{ as follows:}$ $i = 1 + \left( \frac{1}{n} - 1 \right) \alpha$ $\frac{1}{2} = 1 + \left( \frac{1}{n} - 1 \right) \times 1$ $n = 2$

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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
}