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Current Question (ID: 18144)

Question:
$\text{If the relative decrease in vapour pressure is } 0.4 \text{ for a solution containing } 1 \text{ mol NaCl in } 3 \text{ mol of } \text{H}_2\text{O, then percentage ionisation of NaCl will be -}$
Options:
  • 1. $60\%$
  • 2. $80\%$
  • 3. $40\%$
  • 4. $100\%$
Solution:
$\frac{\Delta P}{P_o} = i X_{\text{solute}}$ $\text{Step 1:}$ $\text{Calculate the value of } i \text{ as follows:}$ $\frac{\Delta P}{P_o} = i X_{\text{solute}}$ $0.4 = \frac{i \cdot 1}{1+3}$ $0.4i + 1.2 = i$ $0.4i - i = -1.2$ $-0.6i = -1.2$ $i = 2$ $\text{Step 2:}$ $\text{Calculate the value of } \alpha \text{ as follows:}$ $\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-$ $\frac{1-\alpha}{1} \quad \frac{\alpha}{1} \quad \frac{\alpha}{1}$ $\text{The relation between } i \text{ and } \alpha \text{ is as follows:}$ $i = \frac{\text{Total number of particles after dissociation}}{\text{Total number of particles before dissociation}}$ $i = \frac{1-\alpha+\alpha}{1}$ $2 = 1 + \alpha$ $1 = \alpha$ $\alpha = 100\%$

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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
}