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Current Question (ID: 8430)

Question:

$\text{Pure liquids and solids are ignored while writing the expression for the equilibrium constant because:}$

Options:

1. $\text{Size and shape of a pure substance are always fixed.}$
2. $\text{Volume of solids and liquids is relatively fixed.}$
3. $\text{Charges and masses of pure substances are always fixed.}$
4. $\text{All of the above}$

Solution:

$\text{Hint: Density and molecular mass of a pure substance is always fixed at particular temperature.}$ $\text{Explanation:}$ $\text{For a pure substance (both solids and liquids),}$ $[\text{Pure}] \text{ substance} = \frac{\text{Number of moles}}{\text{Volume}}$ $= \frac{\frac{\text{Mass}}{\text{molecular mass}}}{\text{Volume}}$ $= \frac{\text{Mass}}{\text{Volume} \times \text{Molecular mass}}$ $= \frac{\text{Density}}{\text{Molecular mass}}$ $\text{Now, the molecular mass and density (at a particular temperature) of a pure substance is always fixed and is accounted for in the equilibrium constant.}$ $\text{Therefore, the values of pure substances are not mentioned in the equilibrium constant expression.}$

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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
}