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Current Question (ID: 18052)

Question:

$\text{The Henry's law constant for the solubility of } \text{N}_2 \text{ gas in water at } 298 \text{ K is } 1.0 \times 10^5 \text{ atm.}$ $\text{The mole fraction of } \text{N}_2 \text{ in air is } 0.8. \text{ The number of moles of } \text{N}_2 \text{ formed from air dissolved in } 10 \text{ moles of water at } 298 \text{ K and } 5 \text{ atm pressure is-}$

Options:

1. $4.0 \times 10^{-4} \text{ mol}$
2. $4.0 \times 10^{-5} \text{ mol}$
3. $5.0 \times 10^{-4} \text{ mol}$
4. $4.0 \times 10^{-6} \text{ mol}$

Solution:

$\text{Hint: Henry's law, } P = K_H X_{\text{solute}}$ $\text{Step 1:}$ $\text{The mole fraction of nitrogen in the air is } 0.8 \text{ and the total pressure is } 5 \text{ atm.}$ $\text{Calculate the partial pressure of nitrogen using the Dalton law of partial pressure as follows:}$ $P = P_T X_{\text{N}_2}$ $P = 5 \times 0.8$ $P = 4 \text{ atm}$ $\text{Step 2:}$ $\text{Using Henry's law, calculate the number of moles of nitrogen in the solution as follows:}$ $\text{Mole fraction of } \text{N}_2 = \frac{n_1}{n_1+n_2}$ $K_H = 1 \times 10^5 \text{ atm}$ $X_{\text{N}_2} = \frac{P_{\text{N}_2}}{K_H}$ $= \frac{4}{10^5}$ $X_{\text{N}_2} = 4 \times 10^{-5}$ $4 \times 10^{-5} = \frac{n_1}{10+n_1}$ $4 \times 10^{-5} = \frac{n_1}{10}$ $n_1 = 4 \times 10^{-4}$ $\text{Hence, the number of moles of nitrogen is } 4 \times 10^{-4} \text{ mol.}$

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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
}