Import Question JSON

Current Question (ID: 18122)

Question:

$\text{An aqueous solution containing 1 g of urea boils at } 100.25 \, ^\circ\text{C.}$ $\text{The aqueous solution containing 3 g of glucose in the same volume will boil at:}$

Options:

1. $100.75 \, ^\circ\text{C}$
2. $100.5 \, ^\circ\text{C}$
3. $100 \, ^\circ\text{C}$
4. $100.25 \, ^\circ\text{C}$

Solution:

$\text{Hint: } k_b \text{ value for both the solution is the same.}$ $\text{Step 1:}$ $\text{For urea,}$ $T_b - T_b^\circ = k_b \times \frac{\text{amount of urea}}{\text{molar mass of urea}} \times \frac{1000}{\text{amount of solvent}} \quad \cdots (1)$ $\text{For glucose,}$ $T_b' - T_b^\circ = k_b \times \frac{\text{amount of glucose}}{\text{molar mass of glucose}} \times \frac{1000}{\text{amount of solvent}} \quad \cdots (2)$ $\text{Step 2:}$ $\text{Compare two equations with respect to } k_b$ $\left(T_b' - T_b^\circ\right) \times \frac{\text{molar mass of glucose}}{\text{amount of glucose}} \times \frac{\text{amount of solvent}}{1000} = \left(T_b - T_b^\circ\right) \times \frac{\text{molar mass of urea}}{\text{amount of urea}} \times \frac{\text{amount of solvent}}{1000}$ $\left(T_b' - 100\right) \times \frac{180}{3} = \left((100.25 - 100)\right) \times \frac{60}{1}$ $\left(T_b' - 100\right) \times 60 = (100.25 - 100) \times 60$ $T_b' = 100.25$

Import JSON File

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
}