Import Question JSON

Current Question (ID: 8151)

Question:

$\text{2.9 g of a gas at 95 }^\circ\text{C occupies the same volume as 0.184 g of dihydrogen at 17 }^\circ\text{C, at the same pressure. The molar mass of the gas is -}$

Options:

1. $20\text{ g mol}^{-1}$
2. $40\text{ g mol}^{-1}$
3. $34\text{ g mol}^{-1}$
4. $14\text{ g mol}^{-1}$

Solution:

$\text{Hint: Use Ideal gas equation}\n\n\text{Step 1: From the given data write down the expression of volume for both the gases.}\n\n\text{Volume (V) occupied by dihydrogen is given by,}\n\n\text{V} = \frac{\text{m}}{\text{M}} \frac{\text{RT}}{\text{p}}\n= \frac{0.184}{2} \times \frac{\text{R}\times 290}{\text{p}}\n\n\text{Let M be the molar mass of the unknown gas. Volume (V) occupied by the unknown can be calculated as:}\n\n\text{V} = \frac{\text{m}}{\text{M}} \frac{\text{RT}}{\text{p}}\n= \frac{2.9}{\text{M}} \times \frac{\text{R}\times 368}{\text{p}}\n\n\text{Step 2: Find out the molar mass of unknown gas using the given condition.}\n\n\text{According to the question,}\n\n\frac{0.184}{2} \times \frac{\text{R}\times 290}{\text{p}} = \frac{2.9}{\text{M}} \times \frac{\text{R}\times 368}{\text{p}}\n\Rightarrow \frac{0.184}{2} \times 290 = \frac{2.9 \times 368}{\text{M}}\n\Rightarrow \text{M} = \frac{2.9 \times 368 \times 2}{0.184 \times 290}\n\Rightarrow \text{M} = 40\text{ g mol}^{-1}\n\n\text{Hence, the molar mass of the gas is 40 g mol}^{-1}\text{.}$

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
}