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Current Question (ID: 21398)

Question:

$\text{The mole fraction of glucose (C}_6\text{H}_{12}\text{O}_6\text{) in an aqueous binary solution is 0.1. The mass percentage of water in it to the nearest integer is:}$

Options:

1. $44\%$
2. $49\%$
3. $47\%$
4. $41\%$

Solution:

$\text{Hint: The mass percentage of water formula is } = \frac{\text{Mass of water}}{\text{mass of water} + \text{mass of glucose}} \times 100$ $\text{Step 1:}$ $\text{The mole fraction of glucose in a solution is 0.1. If total mole of the solution is 1. Then mole of water is 0.9.}$ $\text{The mass percentage of water formula is}$ $= \frac{\text{Mass of water}}{\text{mass of water} + \text{mass of glucose}} \times 100$ $\text{Step 2:}$ $\text{Calculate the mass percentage of water is as follows:}$ $\text{mass}\% \text{ of water} = \frac{0.9 \times 18}{0.1 \times 180 + 0.9 \times 18} \times 100$ $= 47.36$ $\text{The mass percentage of water to the nearest integer, is 47 \%}$

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
}