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Current Question (ID: 11121)

Question:

$\text{We have two vessels of equal volume, one filled with hydrogen and the other with equal mass of helium. The common temperature is } 27^{\circ} \text{C. What is the relative number of molecules in the two vessels?}$

Options:

1. $\frac{n_{H}}{n_{He}} = \frac{1}{1}$
2. $\frac{n_{H}}{n_{He}} = \frac{5}{1}$
3. $\frac{n_{H}}{n_{He}} = \frac{2}{1}$ (Correct)
4. $\frac{n_{H}}{n_{He}} = \frac{3}{1}$

Solution:

$\text{Hint: Moles } (n) = \frac{\text{mass}}{\text{molecular mass}}. \text{ The number of molecules is directly proportional to the number of moles. The mass of hydrogen and helium in the two vessels is equal. Let the mass be } m. \text{ The number of moles for hydrogen } (n_H) \text{ is } n_H = \frac{m}{\text{Molar mass of H}_2} = \frac{m}{2} \text{ g/mol. The number of moles for helium } (n_{He}) \text{ is } n_{He} = \frac{m}{\text{Molar mass of He}} = \frac{m}{4} \text{ g/mol. The ratio of the number of moles (and therefore, the number of molecules) is given by } \frac{n_H}{n_{He}} = \frac{\frac{m}{2}}{\frac{m}{4}} = \frac{m}{2} \cdot \frac{4}{m} = \frac{4}{2} = 2. \text{ Therefore, } \frac{n_H}{n_{He}} = \frac{2}{1}. \text{ So, the relative number of molecules is } 2 \text{ to } 1. \text{ The temperature and volume information is extra data and is not needed for the calculation. Hence, option 3 is the correct answer.}$

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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
}