Import Question JSON

Current Question (ID: 11142)

Question:

$\text{If } \text{V}_{\text{H}}, \text{V}_{\text{N}} \text{ and } \text{V}_{\text{O}} \text{ denote the root-mean square velocities of molecules of hydrogen, nitrogen and oxygen respectively at a given temperature, then:}$

Options:

1. $\text{V}_{\text{N}} > \text{V}_{\text{O}} > \text{V}_{\text{H}}$
2. $\text{V}_{\text{H}} > \text{V}_{\text{N}} > \text{V}_{\text{O}}$ (Correct)
3. $\text{V}_{\text{O}} > \text{V}_{\text{N}} > \text{V}_{\text{H}}$
4. $\text{V}_{\text{O}} > \text{V}_{\text{H}} > \text{V}_{\text{N}}$

Solution:

$\text{The root-mean square velocity is given by the formula } \text{v}_{\text{rms}} = \sqrt{\frac{3\text{RT}}{\text{M}}}. \text{ At a given temperature T, 3RT is constant. Hence, the root-mean square velocity is inversely proportional to the square root of the molar mass, i.e., } \text{v}_{\text{rms}} \propto \frac{1}{\sqrt{\text{M}}}. \text{The molar masses of hydrogen } (\text{H}_2), \text{ nitrogen } (\text{N}_2) \text{ and oxygen } (\text{O}_2) \text{ are: M}_{\text{H}_2} = 2 \text{ g/mol}, \text{ M}_{\text{N}_2} = 28 \text{ g/mol}, \text{ and } \text{M}_{\text{O}_2} = 32 \text{ g/mol}. \text{Since the molar mass of hydrogen is the smallest, its root-mean square velocity will be the largest. Conversely, oxygen has the largest molar mass, so its root-mean square velocity will be the smallest. Thus, the order of velocities is } \text{V}_{\text{H}} > \text{V}_{\text{N}} > \text{V}_{\text{O}}. \text{ Therefore, option 2 is correct.}$

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
}