Import Question JSON

Current Question (ID: 8238)

Question:

$\text{The rate of diffusion of methane at a given temperature is twice that of a gas X. The molecular weight of X is}$

Options:

1. $64.0$
2. $32.0$
3. $4.0$
4. $8.0$

Solution:

$r_{\text{CH}_4}/r_X = \sqrt{(M_x/M_{\text{CH}_4})}$ $2 = \sqrt{(M_x/16)}$ $M_x = 64$ $\text{According to Graham's law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molecular weight.}$ $\text{Given that the rate of diffusion of methane (CH}_4\text{) is twice that of gas X, we can write:}$ $\frac{r_{\text{CH}_4}}{r_X} = \sqrt{\frac{M_X}{M_{\text{CH}_4}}}$ $2 = \sqrt{\frac{M_X}{16}}$ $\text{Squaring both sides:}$ $4 = \frac{M_X}{16}$ $M_X = 64$ $\text{Therefore, the molecular weight of gas X is 64.0.}$

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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
}