Import Question JSON

Current Question (ID: 18014)

Question:

$\text{A solution containing } 10 \text{ g/dm}^3 \text{ of urea (molecular mass = } 60 \text{ g mol}^{-1}) \text{ is isotonic with a } 5\% \text{ solution of a non-volatile solute. The molecular mass of this non-volatile solute is:}$

Options:

1. $25 \text{ g mol}^{-1}$
2. $300 \text{ g mol}^{-1}$
3. $350 \text{ g mol}^{-1}$
4. $200 \text{ g mol}^{-1}$

Solution:

$\text{Hint: Isotonic solution has an equal osmotic pressure}$ $\text{Step 1:}$ $\text{Molarity is the number of moles per litre of solution.}$ $\text{Note that } 1 \text{ dm}^3=1 \text{ L}$ $\text{Molarity of urea solution is}$ $= \frac{10}{60} \text{ M} = \frac{1}{6} \text{ M}$ $\text{Step 2:}$ $\text{Also, note that } 5\% \text{ solution of the non-volatile solution means } 5 \text{ g solute in } 100 \text{ ml solution. For } 1 \text{ L or } 1000 \text{ ml solution, } 50 \text{ g of nonvolatile solute will be required.}$ $\text{Molarity of } 5\% \text{ non-volatile solute}$ $= \frac{50}{m} \text{ M}$ $\text{where } m=\text{Mol. Wt. of a non-volatile solute}$ $\text{Step 3:}$ $\text{Now since, both the solutions are isotonic with each other (given)}$ $\frac{1}{6} = \frac{50}{m}$ $\text{Or } m = 50 \times 6 = 300 \text{ g mol}^{-1}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}