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Current Question (ID: 7446)

Question:

$\text{Symbols } ^{79}_{35}\text{Br} \text{ and } ^{79}\text{Br} \text{ can be written, whereas symbols } ^{35}_{79}\text{Br} \text{ and } ^{35}\text{Br} \text{ are not acceptable because:}$

Options:

1. $\text{The general convention for representing an element along with its atomic number (A) and atomic mass (Z) is } ^{A}_{Z}\text{X}$
2. $\text{The general convention for representing an element along with its atomic mass (A) and atomic number (Z) is } ^{A}_{Z}\text{X}$
3. $\text{The general convention for representing an element along with its wavelength (A) and frequency (Z) is } ^{A}_{Z}\text{X}$
4. $\text{The general convention for representing an element along with its isotopes (A) and atomic number (Z) is } ^{A}_{Z}\text{X}$

Solution:

$\text{Hint: See the general convention of representing an element with atomic mass and atomic number.}$ $\text{Explanation:}$ $\text{The general convention of representing an element along with its atomic mass (A) and atomic number (Z) is } ^{A}_{Z}\text{X}$ $\text{Hence, } ^{79}_{35}\text{Br} \text{ is acceptable but } ^{35}_{79}\text{Br} \text{ is not acceptable.}$ $^{79}\text{Br} \text{ can be written but } ^{35}\text{Br} \text{ cannot be written because the atomic number of an element is constant, but the atomic mass of an element depends upon the relative abundance of its isotopes. Hence, it is necessary to mention the atomic mass of an element.}$

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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
}