Import Question JSON

Current Question (ID: 7440)

Question:

$\text{Which of the following is dependent on temperature?}$ $\begin{array}{|c|c|c|c|}\hline\text{1.} & \text{Molarity} & \text{2.} & \text{Mole fraction} \\ \hline\text{3.} & \text{Weight percentage} & \text{4.} & \text{Molality} \\ \hline\end{array}$

Options:

1. $\text{Molarity}$
2. $\text{Mole fraction}$
3. $\text{Weight percentage}$
4. $\text{Molality}$

Solution:

$\text{Hint: The concentration terms depend on volume change with temperature}$ $\text{Explanation:}$ $\text{Molarity depends on the volume of a solution which can be changed with the}$ $\text{change in temperature.}$ $\text{Molarity is defined as the number of moles of solute per liter of solution.}$ $\text{Since the volume of a solution typically expands when heated and contracts}$ $\text{when cooled, the molarity of a solution will change with temperature even}$ $\text{if the amount of solute remains constant.}$ $\text{In contrast, mole fraction, weight percentage, and molality are all independent}$ $\text{of temperature because they are based on mass/moles ratios rather than volume.}$ $\text{Mole fraction is the ratio of moles of one component to the total moles in the solution.}$ $\text{Weight percentage is the ratio of the mass of solute to the total mass of the solution.}$ $\text{Molality is the number of moles of solute per kilogram of solvent.}$ $\text{Therefore, option 1 (Molarity) is the correct answer.}$

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
}