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Current Question (ID: 18073)

Question:

$\text{The vapour pressure of 1 molal solution of a non-volatile solute in water at 300 K would be:}$ $\text{(The vapour pressure of water at 300 K = 12.3 kPa)}$

Options:

1. $21.08 \text{ kPa}$
2. $12.08 \text{ kPa}$
3. $33.08 \text{ kPa}$
4. $4.08 \text{ kPa}$

Solution:

$\text{Hint: Use expression of relative lowering of vapour pressure.}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).}$ $\text{The molar mass of water} = 18 \text{ g mol}^{-1}$ $\text{Number of moles present in 1000 g of water} = 55.56 \text{ mol}$ $\therefore \text{ Number of moles are present in 1000 g of water} = \frac{1000}{18} = 55.56 \text{ mol}$ $\text{Step 2:}$ $\text{Therefore, the mole fraction of the solute in the solution is}$ $x_2 = \frac{1}{1+55.56} = 0.0177$ $\text{It is given that,}$ $\text{Vapour pressure of water, } p_1^0 = 12.3 \text{ kPa}$ $\text{STEP 3: Applying the relation,}$ $\frac{p_1^0 - p_1}{p_1^0} = x_2$ $\Rightarrow \frac{12.3 - p_1}{12.3} = 0.0177$ $\Rightarrow 12.3 - p_1 = 0.2177$ $\Rightarrow p_1 = 12.0823$ $= 12.08 \text{ kPa (approximately)}$ $\text{Hence, the vapour pressure of the solution is 12.08 kPa.}$

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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
}