Import Question JSON

Current Question (ID: 10791)

Question:

$\text{A temperature of } 100°\text{F (Fahrenheit scale) is equal to } T \text{ K (Kelvin scale). The value of } T \text{ is:}$

Options:

1. $310.9$ (Correct)
2. $37.8$
3. $100$
4. $122.4$

Solution:

$\text{Hint: } \frac{F - 32}{9} = \frac{C}{5}$ $\text{Step: Find the value of } T \text{ in Kelvin.}$ $\frac{F - 32}{9} = \frac{C}{5}$ $\Rightarrow \frac{100 - 32}{9} = \frac{C}{5} \Rightarrow C = \frac{68 \times 5}{9} = 37.7°\text{C}$ $\text{The value of temperature in Kelvin is given by;}$ $C = 37.7 + 273 = 310.7 \approx 310.9 \text{ K}$ $\text{Therefore, the value of the } T \text{ is } 310.9.$ $\text{Hence, option (1) is the correct answer.}$

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
}