Import Question JSON

Current Question (ID: 18027)

Question:

$\text{Match the terms given in Column I with expressions given in Column II.}$ $\begin{array}{|c|c|} \hline \text{Column I (Term)} & \text{Column II (Expression)} \\ \hline \text{A. Mass percentage} & \frac{\text{Mass of the solute component in solution}}{\text{Total Mass of solution}} \times 100 \\ \text{B. Volume percentage} & \frac{\text{Volume of the solute component in solution}}{\text{Total volume of solution}} \times 100 \\ \text{C. Molarity} & \frac{\text{Number of moles of the solute component}}{\text{Volume of solution in litres}} \\ \text{D. Molality} & \frac{\text{Number of moles of the solute component}}{\text{Mass of solvent in kilograms}} \\ \hline \end{array}$

Options:

1. $1 \ 2 \ 3 \ 4$
2. $2 \ 1 \ 3 \ 4$
3. $1 \ 4 \ 3 \ 2$
4. $4 \ 3 \ 1 \ 2$

Solution:

$\text{Hint: Molarity is defined as the number of moles of solute per litre of solution (moles/Liter)}$ $\begin{array}{|c|c|} \hline \text{Column I} & \text{Column II} \\ \hline \text{(Concentration terms)} & \text{(Mathematical formula)} \\ \hline \text{A. Mass percentage} & \frac{\text{Mass of the solute component in solution}}{\text{Total mass of the solution}} \times 100 \\ \text{B. Volume percentage} & \frac{\text{Volume of the solute component in solution}}{\text{Total volume of solution}} \times 100 \\ \text{C. Molality} & \frac{\text{Number of moles of the solute component}}{\text{Mass of solvent in kilograms}} \\ \text{D. Molarity} & \frac{\text{Number of moles of the solute component}}{\text{Volume of solution in litres}} \\ \hline \end{array}$

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
}