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Current Question (ID: 19849)

Question:

$\text{A mixture of hydrogen and oxygen has a volume of } 2000 \ \text{cm}^3, \text{ temperature } 300 \ \text{K}, \text{ pressure } 100 \ \text{kPa and mass } 0.76 \ \text{g.}$ $\text{The ratio of the number of moles of hydrogen to the number of moles of oxygen in the mixture will be:}$

Options:

1. $\frac{1}{3}$
2. $3 \colon 1$
3. $\frac{1}{16}$
4. $16 \colon 1$

Solution:

$\text{Hint: } PV = (n_1 + n_2)RT$ $\text{Step 1: Find the relation between } n_1 \text{ and } n_2 \text{ from given mass of the mixture.}$ $\text{Let the number of moles of hydrogen and oxygen in the given mixture are } n_1 \text{ and } n_2 \text{ respectively.}$ $\text{Given that the mass of the mixture is } 0.76 \ \text{g.}$ $2n_1 + 32n_2 = 0.76$ $\Rightarrow n_1 + 16n_2 = 0.38 \ldots (1)$ $\text{Step 2: Find the number of moles of hydrogen and oxygen in the given mixture.}$ $\text{Apply the ideal gas equation to the given mixture,}$ $100 \times 10^3 \times 2000 \times 10^{-6} = (n_1 + n_2) \times \frac{25}{3} \times 300 \ldots (2)$ $\text{From (1) and (2) we get;}$ $\frac{2}{25} = 0.38 - 16n_2 + n_2$ $0.08 = 0.38 - 15n_2$ $15n_2 = 0.3$ $\Rightarrow n_2 = 0.02$ $\Rightarrow n_1 = 0.06$ $\text{Step 3: The ratio of the number of moles.}$ $\frac{n_1}{n_2} = \frac{0.06}{0.02} = \frac{3}{1}$ $\text{Therefore, the ratio of the number of moles of hydrogen to the number of moles of oxygen in the mixture is } 3 \colon 1$ $\text{Hence, option (2) is the correct answer.}$

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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
}