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Current Question (ID: 21396)

Question:

$\text{If a compound AB dissociates to the extent of 75\% in an aqueous solution, the molality of the solution which shows a 2.5 K rise in the boiling point of the solution is:}$ $\text{(Rounded-off to the nearest integer) } [K_b = 0.52 \text{ K kg mol}^{-1}]$

Options:

1. 1 (Correct)
2. 5
3. 7
4. 2

Solution:

$\text{Hint: Use the formula of elevation in boiling point, that is, } \Delta T_b = iK_b m$ $\text{Step 1:}$ $\text{The formula of elevation in boiling point is as follows:}$ $\Delta T_b = iK_b m$ $\text{First, calculate the value of } i \text{ using the value of } \alpha.$ $\text{The relation between } i \text{ and } \alpha \text{ for AB is as follows:}$ $i = 1 + \alpha$ $i = 1 + 0.75; = 1.75$ $\text{Step 2:}$ $\text{Calculate the value of molality as follows:}$ $\text{The given values are as follows:}$ $\Delta T_b = 2.5 \text{ K}$ $i = 1.75$ $K_f = 0.52 \text{ K kg mol}^{-1}$ $\Delta T_b = iK_b m$ $2.5 = 1.75 \times 0.52 \times m$ $m = \frac{2.5}{1.75 \times 0.52} = 2.74$ $\therefore \text{ nearest integer answer will be } 3$

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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
}