Import Question JSON

Current Question (ID: 19881)

Question:

$\text{The correct order of root mean square speed, } v_{\text{rms}} \text{ for Ne, Cl}_2 \text{ and OF}_6 \text{ at the same temperature is:}$ $1.\ (v_{\text{rms}})_{\text{Ne}} < (v_{\text{rms}})_{\text{Cl}_2} < (v_{\text{rms}})_{\text{OF}_6}$ $2.\ (v_{\text{rms}})_{\text{Cl}_2} < (v_{\text{rms}})_{\text{Ne}} < (v_{\text{rms}})_{\text{OF}_6}$ $3.\ (v_{\text{rms}})_{\text{OF}_6} < (v_{\text{rms}})_{\text{Cl}_2} < (v_{\text{rms}})_{\text{Ne}}$ $4.\ (v_{\text{rms}})_{\text{OF}_6} < (v_{\text{rms}})_{\text{Ne}} < (v_{\text{rms}})_{\text{Cl}_2}$

Options:

1. $(v_{\text{rms}})_{\text{Ne}} < (v_{\text{rms}})_{\text{Cl}_2} < (v_{\text{rms}})_{\text{OF}_6}$
2. $(v_{\text{rms}})_{\text{Cl}_2} < (v_{\text{rms}})_{\text{Ne}} < (v_{\text{rms}})_{\text{OF}_6}$
3. $(v_{\text{rms}})_{\text{OF}_6} < (v_{\text{rms}})_{\text{Cl}_2} < (v_{\text{rms}})_{\text{Ne}}$
4. $(v_{\text{rms}})_{\text{OF}_6} < (v_{\text{rms}})_{\text{Ne}} < (v_{\text{rms}})_{\text{Cl}_2}$

Solution:

$\text{Hint: } v_{\text{rms}} \propto \frac{1}{\sqrt{M}}$ $\text{Step: Find the correct order of root mean square speed for the given gases.}$ $\therefore v_{\text{rms}} \propto \frac{1}{\sqrt{M}}$ $\text{also, } M_{\text{OF}_6} > M_{\text{Cl}_2} > M_{\text{Ne}}$ $\therefore (v_{\text{rms}})_{\text{OF}_6} < (v_{\text{rms}})_{\text{Cl}_2} < (v_{\text{rms}})_{\text{Ne}}$ $\text{Hence, option (3) is the correct answer.}$

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
}