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Current Question (ID: 18100)

Question:
$\text{When 20 g of naphthoic acid (C}_{11}\text{H}_{8}\text{O}_{2}) \text{ is dissolved in 50 g of benzene, a freezing point depression of 2 K is observed. What is the Van't Hoff factor (i)?}$ $\left(K_f = 1.72 \text{ K kg mol}^{-1}\right)$
Options:
  • 1. $0.5$
  • 2. $1$
  • 3. $2$
  • 4. $3$
Solution:
$\text{Hint: } i = \frac{\text{observed value of colligative property}}{\text{theoretical value of colligative property}}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{The molecular weight of naphthoic acid}$ $\text{C}_{11}\text{H}_{8}\text{O}_{2} = 172 \text{ g/mol}$ $\text{The theoretical value of depression in freezing point}$ $= K_f \times \text{molality} = 1.72 \times \frac{20 \times 1000}{172 \times 50} = 4 \text{ K}$ $\text{Step 2:}$ $\text{Van't Hoff factor}$ $i = \frac{\text{Observed colligative property}}{\text{Calculated colligative property}} = \frac{2}{4} = 0.5$

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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
}