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Current Question (ID: 8236)

Question:
$\text{Density ratio of O}_2 \text{ and H}_2 \text{ is 16 : 1. The ratio of their r.m.s. velocities will be-}$
Options:
  • 1. $4 : 1$
  • 2. $1 : 16$
  • 3. $1 : 4$
  • 4. $16 : 1$
Solution:
$\text{Hint: Root-mean square (RMS) velocity concept}$ $\text{Explanation:}$ $\text{Density ratio of O}_2 \text{ and H}_2 \text{ is 16:1. The ratio of their RMS velocities will be 1:4.}$ $\text{This is because rms speed is inversely proportional to the square root of density.}$ $v_{\text{rms}} = \sqrt{\frac{3 \text{RT}}{\text{m}}}$ $\propto \frac{1}{\text{M}}$ $\frac{C_{\text{rms}}\text{ O}_2}{C_{\text{rms}}\text{ H}_2} = \sqrt{\frac{\text{M}_{\text{H}_2}}{\text{M}_{\text{O}_2}}} = \sqrt{\frac{\text{d}_{\text{H}_2}}{\text{d}_{\text{O}_2}}} = \sqrt{\frac{1}{16}} = 1 : 4$

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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
}