Import Question JSON

Current Question (ID: 8130)

Question:

\text{If } 1 \text{ g of each of the following gases are taken at STP, then the gas that will have the highest volume is - } \text{CO, H}_2\text{O, CH}_4\text{, NO}

Options:

1. $\text{CH}_4$
2. $\text{H}_2\text{O}$
3. $\text{NO}$
4. $\text{CO}$

Solution:

$\text{Hint: Avogadro's gas law.} \text{Explanation:} \text{Step 1:} \text{From Avogadro's law, we know that} \text{The volume of 1 mole of the gas = gram molecular mass = } 22.4 \text{ L at STP} \text{Step 2:} \text{Calculate the volume occupied by the following gas as follows:} \text{The volume occupied by } 28 \text{ g CO (1 mol CO) = } 22.4 \text{ L at STP } (\therefore \text{ Molar mass of CO = } 12 + 16 = 28 \text{ g mol}^{-1}) \therefore \text{ The volume occupied by } 1 \text{ g CO} = \frac{22.4}{28} \text{ L at STP} \text{Similarly, the volume occupied by } 1 \text{ g H}_2\text{O} (\therefore \text{ Molar mass of H}_2\text{O = } (2 \times 1) + 16 = 18 \text{ g mol}^{-1}) \text{The volume occupied by } 1 \text{ g CH}_4 = \frac{22.4}{16} \text{ L at STP} (\therefore \text{ Molar mass of CH}_4 = 12 + (4 \times 1) = 16 \text{ g mol}^{-1}) \text{The volume occupied by } 1 \text{ g NO} = \frac{22.4}{30} \text{ L at STP} (\therefore \text{ Molar mass of NO = } 14 + 16 = 30 \text{ g mol}^{-1}) \text{Thus, } 1 \text{ g CH}_4 \text{ will occupy maximum volume while } 1 \text{ g of NO will occupy minimum volume at STP.}$

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
}