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Current Question (ID: 18091)

Question:

$\text{At } 100 \, ^\circ\text{C the vapour pressure of a solution of } 6.5 \, \text{g of a solute in } 100 \, \text{g water is } 732 \, \text{mm. If } K_b = 0.52, \text{ the boiling point of this solution will be:}$

Options:

1. $100 \, ^\circ\text{C}$
2. $102 \, ^\circ\text{C}$
3. $103 \, ^\circ\text{C}$
4. $101 \, ^\circ\text{C}$

Solution:

$\text{Hint: Using relative lowering in the vapor pressure formula, calculate the molar mass of the solute.}$ $\text{Step 1:}$ $\text{The formula for the relative lowering of vapor pressure is as follows:}$ $\frac{P_A - P_B}{P_A} = \frac{n_B}{n_A}$ $\text{Here, } P_A \text{ pure vapour pressure of the solvent, and } P_B \text{ = vapour pressure of the solution}$ $n_B = \text{number of mole of solute and } n_A = \text{number of mole of solvent}$ $\frac{P_A - P_B}{P_A} = \frac{n_B}{n_A}$ $\frac{760 - 732}{760} = \frac{W_B \times M_A}{M_B \times W_A}$ $\frac{28}{760} = \frac{6.5 \times 18}{M_B \times 100}$ $M_B = 31.6 \, \text{g mol}^{-1}$ $\text{Step 2:}$ $\text{The formula for the elevation of the boiling point is as follows:}$ $\Delta T_b = k_b m$ $\text{Here, } k_b \text{ is the boiling point elevation constant, and } m \text{ is molality.}$ $\text{Therefore, } \Delta T_b = 0.52 \times \frac{6.5 \times 1000}{31.6 \times 100} = 1.07.$ $\text{Therefore Boiling point } = 100 + 1.07 = 101 \cdot 07^\circ\text{C} \approx 101^\circ\text{C}$

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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
}