Import Question JSON

Current Question (ID: 10723)

Question:

$\text{The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. The value of these temperatures on Fahrenheit scale will be:}$

Options:

1. $-415.44^\circ \text{F}, -69.88^\circ \text{F}$ (Correct)
2. $-248.58^\circ \text{F}, -56.60^\circ \text{F}$
3. $315.44^\circ \text{F}, -69.88^\circ \text{F}$
4. $415.44^\circ \text{F}, -79.88^\circ \text{F}$

Solution:

$\text{Hint: Use the concept of temperature scales.}$ $\text{Step 1: Find the triple-point temperature of the neon on the Fahrenheit scale.}$ $\text{The conversion formulas are: } t_C = t_K - 273.15 \text{ and } t_F = \frac{9}{5}t_C + 32$ $\text{Given: Triple point of neon: } T_N = 24.57 \text{ K}$ $\text{Convert to Celsius:}$ $T_{N,C} = T_N - 273.15 = 24.57 - 273.15 = -248.58^\circ \text{C}$ $\text{Convert to Fahrenheit:}$ $T_{N,F} = (-248.58) \times \frac{9}{5} + 32 = -415.44^\circ \text{F}$ $\text{Step 2: Find the triple-point temperature of the carbon dioxide on the Fahrenheit scale.}$ $\text{Given: Triple point of carbon dioxide: } T_{CO_2} = 216.55 \text{ K}$ $\text{Convert to Celsius:}$ $T_{CO_2,C} = T_{CO_2} - 273.15 = 216.55 - 273.15 = -56.60^\circ \text{C}$ $\text{Convert to Fahrenheit:}$ $T_{CO_2,F} = (-56.60) \times \frac{9}{5} + 32 = -69.88^\circ \text{F}$ $\text{Hence, option (1) is the correct answer.}$

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
}