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Current Question (ID: 21399)

Question:

$\text{When 9.45 g of ClCH}_2\text{COOH is added to 500 mL of water, its freezing point drops by 0.5 } ^\circ \text{C.}$ $\text{The dissociation constant of ClCH}_2\text{COOH is } x \times 10^{-3}. \text{ The value of } x \text{ is:}$ $\text{(Rounded off to the nearest integer)}$ $\left[ K_f(\text{H}_2\text{O}) = 1.86 \text{ kgmol}^{-1} \right]$

Options:

1. $36 \times 10^{-3}$
2. $29 \times 10^{-3}$
3. $45 \times 10^{-3}$
4. $30 \times 10^{-3}$

Solution:

$\text{Hint: The formula of depression in freezing point is } \Delta T = iK_f m$ $\text{Step 1:}$ $\text{Calculate the value of } i \text{ as follows:}$ $\Delta T = iK_f m$ $T_f^0 - T_f = iK_f m$ $0.5 = i \times 1.86 \times \frac{9.45}{94.5} \times \frac{1000}{500}$ $0.5 = i \times 0.37$ $i = 1.35$ $\text{Step 2:}$ $\text{Calculate the value of the degree of dissociation.}$ $\text{The reaction is as follows:}$ $\text{ClCH}_2\text{COOH} \rightleftharpoons \text{ClCH}_2\text{COO}^- + \text{H}^+$ $i = 1 + (2 - 1)\alpha$ $i = 1 + \alpha$ $1.35 = 1 + \alpha$ $\alpha = 0.35$ $\text{Step 3:}$ $\text{Calculate the value of the ionisation constant as follows:}$ $\text{ClCH}_2\text{COOH} \rightleftharpoons \text{ClCH}_2\text{COO}^-$ $C - C\alpha \hspace{1cm} C\alpha$ $K_a = \frac{(C\alpha)^2}{C - C\alpha} = \frac{C\alpha^2}{1 - \alpha} ; C = \frac{0.1}{500/1000} = 0.2$ $K_a = \frac{0.2 \times (32/93)^2}{1 - (32/93)} = \frac{0.2 \times (32)^2}{93 \times 61} = 0.036$ $K_a = 36 \times 10^{-3}$

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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
}