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Current Question (ID: 21403)

Question:

$\text{A set of solutions is prepared using 180 g of water as a solvent and 10 g of different non-volatile solutes A, B and C. The relative lowering of vapour pressure in the presence of these solutes are in the order:}$ $\text{[Given, molar mass of A = 100 g mol}^{-1}\text{; B = 200 g mol}^{-1}\text{; C = 10,000 g mol}^{-1}]}$

Options:

1. $\text{A > B > C}$
2. $\text{A > C > B}$
3. $\text{C > B > A}$
4. $\text{B > C > A}$

Solution:

$\text{Hint: Use the formula of relative lowering of vapour pressure}$ $\frac{\Delta P}{P_o} = X_{\text{solute}}$ $\text{Calculate the mole fraction value for each solution because relative lowering in vapour pressure directly proportional to the mole fraction of solute.}$ $X = \frac{\text{number of moles of solute}}{\text{number of moles of solute} + \text{number of moles solvent}} = \frac{\frac{\text{amount of solute}}{\text{molar mass of solute}}}{\frac{\text{amount of solute}}{\text{molar mass of solute}} + \frac{\text{amount of solvent}}{\text{molar mass of solvent}}}$ $\text{Calculate the value of relative lowering of vapour pressure as follow:}$ $\text{For first solution,}$ $\frac{\Delta P}{P_o} = \frac{\frac{10}{100}}{\frac{10}{100} + \frac{180}{18}} = 0.01$ $\text{For second solution,}$ $\frac{\Delta P}{P_o} = 0.0050$ $\text{For third solution,}$ $\frac{\Delta P}{P_o} = 10^{-4}$ $\text{The order of relative lowering of vapour pressure for the given solution is A > B > C}$

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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
}