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Current Question (ID: 21427)

Question:
$\text{A solution containing 2 g of a non-electrolyte solute dissolved in 20 g of water has a boiling point of 373.52 K. Calculate the molecular mass of the solute.}$ $\text{(Given that the ebullioscopic constant (} K_b \text{) is 0.52 K} \cdot \text{kg/mol)}$
Options:
  • 1. $140 \text{ g/mol}$
  • 2. $80 \text{ g/mol}$
  • 3. $120 \text{ g/mol}$
  • 4. $100 \text{ g/mol}$
Solution:
$\text{Hint: } \Delta T_b = K_b \cdot m$ $\text{Step 1: The formula for elevation in boiling point is } \Delta T_b = K_b \cdot m , \text{ where } m \text{ is the molality of the solution.}$ $\text{The given mass of solute = 2 g}$ $\text{Let the molar mass of solute = } M$ $\text{Also, the mass of solution = 20 g = 0.02 kg}$ $\text{The elevation in boiling point = } 373.52 \text{ K} - 373 \text{ K} = 0.52 \text{ K}$ $\text{Step 2: Hence, } \Delta T_b = K_b \cdot m$ $0.52 = 0.52 \times \frac{2/M}{0.02} \text{ (M indicates the molecular mass of solute)}$ $M = 100 \text{ g}$ $\text{Hence, option 4 is the correct answer.}$

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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
}