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Current Question (ID: 21411)

Question:

$18 \text{ g glucose } (\text{C}_6\text{H}_{12}\text{O}_6) \text{ is added to } 178.2 \text{ g water. The vapour pressure of aqueous solution in torr is:}$ $1. \ 7.6$ $2. \ 752.4$ $3. \ 76.0$ $4. \ 780.0$

Options:

1. $7.6$
2. $752.4$
3. $76.0$
4. $780.0$

Solution:

$\text{Hint: Use the formula of relative lowering of vapour pressure}$ $\text{The formula of relative lowering in vapour pressure is as follows:}$ $\frac{\Delta P}{P^o} = X_{\text{solute}}$ $\frac{P^o - P_s}{P^o} = X_{\text{solute}}$ $\frac{P^o - P_s}{P^o} = \frac{\text{number of moles of solute}}{\text{number of moles of solute} + \text{number of moles of solvent}}$ $= \frac{\frac{\text{amount of solute}}{\text{molar mass of solute}}}{\frac{\text{amount of solute}}{\text{molar mass of solute}} + \frac{\text{amount of solvent}}{\text{molar mass of solvent}}}$ $\text{Step 2:}$ $\text{Pressure of pure water} = 760 \text{ torr}$ $\text{Calculate the pressure of the solution as follows:}$ $\frac{760 - P_s}{760} = \frac{\frac{18}{180}}{\frac{18}{180} + \frac{178.2}{18}}$ $P_s = 752.4 \text{ torr}$

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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
}