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Current Question (ID: 18054)

Question:

$\text{An aqueous solution of } 2\% \text{ (w/w) non-volatile solute exerts a pressure of } 1.004 \text{ bar at the normal boiling point of the solvent. Molar mass of the solute would be:}$

Options:

1. $23.69 \text{ g mol}^{-1}$
2. $41.35 \text{ g mol}^{-1}$
3. $56.23 \text{ g mol}^{-1}$
4. $22.76 \text{ g mol}^{-1}$

Solution:

$\text{Hint: Use Raoult's law.}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{Vapour pressure of the solution at normal boiling point } (p_1) = 1.004 \text{ bar}$ $\text{Vapour pressure of pure water at normal boiling point } (p_1^0) = 1.013 \text{ bar}$ $\text{Molar mass of solute } (w_2) = 2 \text{ g}$ $\text{Molar mass of solvent } (w_1) = 98 \text{ g}$ $\text{Step 2:}$ $\text{According to Raoult's law,}$ $\frac{p_1^0 - p_1}{p_1^0} = \frac{w_2 \times M_1}{M_2 \times w_1}$ $\Rightarrow \frac{1.013 - 1.004}{1.013} = \frac{2 \times 18}{M_2 \times 98}$ $\Rightarrow \frac{0.009}{1.013} = \frac{2 \times 18}{M_2 \times 98}$ $\Rightarrow M_2 = \frac{1.013 \times 2 \times 18}{0.009 \times 98}$ $= 41.35 \text{ g mol}^{-1}$

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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
}