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Current Question (ID: 7382)

Question:
$\text{The molar mass of naturally occurring Argon isotopes is:}$ $\begin{array}{|c|c|c|} \hline \text{Isotope} & \text{Isotopic molar mass} & \text{Abundance} \\ \hline \text{36-Ar} & 35.96755 \text{ g mol}^{-1} & 0.337\% \\ \hline \text{38-Ar} & 37.96272 \text{ g mol}^{-1} & 0.063\% \\ \hline \text{40-Ar} & 39.9624 \text{ g mol}^{-1} & 99.600\% \\ \hline \end{array}$
Options:
  • 1. $49.99947 \text{ g mol}^{-1}$
  • 2. $39.99947 \text{ g mol}^{-1}$
  • 3. $35.59947 \text{ g mol}^{-1}$
  • 4. $45.59947 \text{ g mol}^{-1}$
Solution:
$\text{HINT: } \Sigma \text{ natural abundance } \times \text{ isotopic molar mass}$ $\text{Average molar mass is the summation of product of abundance and molar mass of the isotope}$ $\text{Thus, molar mass of naturally occuring Argon is:}$ $= [(35.96755)] \times \frac{0.337}{100} + (37.96272) \times \frac{0.063}{100} + (39.9624) \times \frac{99.600}{100} \text{ g mol}^{-1}$ $= [0.121] + 0.024 + 39.9802 \text{ g mol}^{-1}$ $= 39.9947 \text{ g mol}^{-1}$

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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
}