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Current Question (ID: 21408)

Question:
$\text{At room temperature, a dilute solution of urea is prepared by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be:}$ $\text{(molar mass of urea = 60 g mol}^{-1}\text{)}$
Options:
  • 1. $0.031 \text{ mmHg}$
  • 2. $0.017 \text{ mmHg}$
  • 3. $0.028 \text{ mmHg}$
  • 4. $0.027 \text{ mmHg}$
Solution:
$\text{Hint: } \frac{\Delta p}{p^0} = \frac{n}{N+n}$ $\text{Step 1:}$ $\text{The formula of relative lowering in vapour pressure is as follows:}$ $\frac{\Delta p}{p^0} = \frac{n}{N+n}$ $\text{The given value is as follows:}$ $p^0 = 35 \text{ mm Hg}$ $\text{amount of urea} = 0.60 \text{ g}$ $\text{amount of water} = 360 \text{ g}$ $\text{Step 2:}$ $\text{Calculate the lowering of pressure as follows:}$ $\text{i. e. } \frac{\Delta p}{p^0} = \frac{n}{N+n}$ $\text{or } \Delta p = p^0 \times \frac{n}{(N+n)}$ $\Delta p = 35 \times \frac{0.6}{\frac{360}{18} + \frac{0.6}{60}}$ $\Delta p = \frac{35 \times 06}{600}$ $= \frac{35 \times 100}{2001 \times 100}$ $= 0.017 \text{ mmHg}$ $\text{Hence, option second is the correct answer.}$

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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
}