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Current Question (ID: 18084)

Question:

$\text{The mass of a non-volatile solute (molar mass 40 g mol}^{-1}\text{) that should be dissolved in 114 g octane to reduce its vapour pressure to 80 \% would be:}$

Options:

1. $6 \text{ g}$
2. $7 \text{ g}$
3. $8 \text{ g}$
4. $10 \text{ g}$

Solution:

$\text{HINT: The relative lowering of vapor pressure of a solution having non-volatile solutes is equal to the mole fraction of the solute.}$ $\text{Let the vapour pressure of pure octane be } P_1^0$ $\text{Then, the vapour pressure of the octane after dissolving the non-volatile solute is } \frac{80}{100} P_1^0 = 0.8 P_1^0$ $\text{Molar mass of solute, } M_2 = 40 \text{ g mol}^{-1}$ $\text{Mass of octane, } w_1 = 114 \text{ g}$ $\text{Molar mass of octane, (C}_8\text{H}_{18}), M_1 = 8 \times 12 + 18 \times 1 = 114 \text{ g mol}^{-1}$ $\frac{114 \text{ g}}{114 \text{ g mol}^{-1}} = 1 \text{ mole octane}$ $\text{Now, } \frac{P^0 - P_s}{P^0} = x_2$ $\therefore \frac{P^0 - 0.8 P^0}{P^0} = \frac{(w/40)}{[(w/40)+1]}$ $0.2 \times \left[ \frac{w}{40} + 1 \right] = \frac{w}{40}$ $\text{on solving above, we get}$ $\frac{0.8}{w/40} = 0.2$ $w = 10 \text{ g}$

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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
}