Import Question JSON

Current Question (ID: 18033)

Question:

$\text{What mass of urea is needed to prepare 2.5 kg of a 0.25 m aqueous solution?}$

Options:

1. $73 \text{ g}$
2. $37 \text{ g}$
3. $48 \text{ g}$
4. $24 \text{ g}$

Solution:

$\text{Molar mass of urea} = 60 \text{ g mol}^{-1}$ $0.25 \text{ m urea solution means in 1000 g of water 0.25 moles (0.25 \times 60 = 15 \text{ g}) of urea is present.}$ $\text{i.e. 1000 + 15 g of solution contains 15 g urea.}$ $\text{So, 2.5 kg i.e. 2500 g of solution will contain } \frac{15}{1000} \times 2500 + 15 = 36.95 \approx 37 \text{ g of urea}$

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
}