Import Question JSON

Current Question (ID: 7444)

Question:

\text{Match the following physical quantities with units:} \text{A. Molarity - 2. mol/L} \text{B. Molality - 1. mol/kg} \text{C. Pressure - 4. Pascal} \text{D. Luminous intensity - 3. Candela}

Options:

1. $\text{A-1, B-4, C-2, D-3}$
2. $\text{A-2, B-1, C-4, D-3}$
3. $\text{A-1, B-4, C-3, D-2}$
4. $\text{A-4, B-1, C-3, D-2}$

Solution:

$\text{A. Molarity = concentration in mol L}^{-1}$ $\text{Molarity} = \frac{\text{Number of moles}}{\text{Volume in litres}}$ $\text{B. Molality = concentration in mole per kg solvent}$ $\text{Molality} = \frac{\text{Number of moles}}{\text{Mass of solvent (kg)}}$ $\text{C. The SI unit for pressure is the pascal (Pa), equal to one newton per square}$ $\text{metre (N/m}^2\text{ or kg·m}^{-1}\text{·s}^{-2}\text{). This special}$ $\text{name for the unit was added in 1971; before that, pressure in SI was}$ $\text{expressed simply as N/m}^2\text{.}$ $\text{D. Unit of luminous intensity = candela}$ $\text{The candela is the luminous intensity, in a given direction, of a source that}$ $\text{emits monochromatic radiation of frequency 540 × 10}^{12}\text{ hertz and that has}$ $\text{a radiant intensity in that direction of 1/683 watt per steradian.}$ $\text{Therefore, the correct answer is option 2.}$

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
}