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Current Question (ID: 18143)

Question:

$0.5$ $\text{molal aqueous solution of a weak acid (HX) is } 20\% \text{ ionised. The lowering in the freezing point of the solution will be:}$ $[K_f \text{ for water } = 1.86 \text{ K kg mol}^{-1}]$

Options:

1. $-1.12 \text{ K}$
2. $0.56 \text{ K}$
3. $1.12 \text{ K}$
4. $-0.56 \text{ K}$

Solution:

$\text{Hint: } \Delta T_f = i \times K_f \times m$ $\text{Step 1:}$ $\text{The reaction is as follows:}$ $\text{HX} \rightleftharpoons \text{H}^+ + \text{X}^-$ $\text{Weak acid}$ $t=0 \quad 1 \quad 0 \quad 0$ $\text{at eq } 1-\alpha \quad \alpha \quad \alpha$ $\alpha = 20\% \text{ dissociation (given)}$ $i.e., \alpha = 0.2$ $\text{Step 2:}$ $\text{Van't Hoff factor } (i) \text{ can be calculated as:}$ $i = 1 - \alpha + \alpha + \alpha$ $= 1 + \alpha = 1 + 0.2 = 1.2$ $\text{Since } \Delta T_f = i \times K_f \times m$ $= 1.2 \times 1.86 \text{ K kg mol}^{-1} \times 0.5$ $= 1.12 \text{ K}$

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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
}