Import Question JSON

Current Question (ID: 7472)

Question:

$\text{The total number and mass of neutrons in 7 mg of }^{14}\text{C would be:}$ $\text{(Assume that mass of a neutron} = 1.675 \times 10^{-27} \text{ kg)}$

Options:

1. $2.41 \times 10^{21}, 4.03 \times 10^{-6} \text{ kg}$
2. $6.23 \times 10^{23}, 1.67 \times 10^{-21} \text{ kg}$
3. $1.22 \times 10^{22}, 4.03 \times 10^{6} \text{ kg}$
4. $2.41 \times 10^{21}, 4.03 \times 10^{-6} \text{ g}$

Solution:

$\text{HINT: Use mole concept}$ $\text{(i) STEP 1: Find out the number of atoms in 1 mole and the number of neutrons in 1 mole.}$ $\text{Number of atoms of }^{14}\text{C in 1 mole} = 6.023 \times 10^{23}$ $\text{Since 1 atom of }^{14}\text{C contains (14 - 6) i.e., 8 neutrons, the number of neutrons in 14 g of }^{14}\text{C is } (6.023 \times 10^{23}) \times 8. \text{ Or, 14 g of }^{14}\text{C contains } (6.022 \times 10^{23} \times 8) \text{ neutrons.}$ $\text{STEP 2: Find out the number of neutrons in given amount}$ $\text{Number of neutrons in 7 mg}$ $= \frac{6.023}{1400} \times 10^{23} \times 8 \times 7 \text{ mg} = 2.41 \times 10^{21}$ $\text{(ii) STEP 1: Write down the mass one neutron}$ $\text{Mass of one neutron} = 1.67 \times 10^{-27} \text{ kg}$ $\text{STEP 2: Calculate the mass of total calculated neutrons}$ $\text{Mass of total neutrons in 7 g of }^{14}\text{C}$ $= (2.41 \times 10^{21})(1.67 \times 10^{-27} \text{ kg}) = 4.03 \times 10^{-6} \text{ kg}$

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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
}