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Current Question (ID: 18053)

Question:

$\text{The solubility of } \text{H}_2\text{S} \text{ in water at STP is } 0.195 \text{ m. The value of Henry's constant is-}$

Options:

1. $274 \text{ atm}$
2. $285 \text{ atm}$
3. $295 \text{ atm}$
4. $278 \text{ atm}$

Solution:

$\text{Hint: Use the formula of Henry's law}$ $\text{Step 1:}$ $\text{The formula of Henry's law is as follows:}$ $p = K_H X$ $\text{Calculate the mole fraction of } \text{H}_2\text{S} \text{ as follows:}$ $\text{It is given that the solubility of } \text{H}_2\text{S} \text{ in water at STP is } 0.195 \text{ m, i.e., } 0.195 \text{ mol of } \text{H}_2\text{S} \text{ is dissolved in } 1000 \text{ g of water.}$ $\text{Moles of water } = \frac{1000 \text{ g}}{18 \text{ g mol}^{-1}} = 55.56 \text{ mol}$ $\text{Mole of } \text{H}_2\text{S} = 0.195 \text{ mol}$ $\therefore \text{Mole fraction of } \text{H}_2\text{S}, X = \frac{\text{Moles of } \text{H}_2\text{S}}{\text{Moles of } \text{H}_2\text{S} + \text{Moles of water}}$ $X = \frac{0.195}{0.195 + 55.56}$ $X = 0.0035$ $\text{Step 2:}$ $\text{At STP, pressure (p) = 1 atm}$ $\text{According to Henry's law: } p = K_H X$ $K_H = \frac{p}{X}$ $= \frac{1}{0.0035}$ $= 285 \text{ atm}$

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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
}