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Current Question (ID: 7336)

Question:

$\text{Which of the following statement indicates that the law of multiple proportions is being followed?}$

Options:

1. $\text{A sample of carbon dioxide taken from any source will always have carbon and oxygen in the ratio of } 1:2\text{.}$
2. $\text{Carbon forms two oxides namely, } \text{CO}_2 \text{ and } \text{CO} \text{ where masses of oxygen that combine with a fixed mass of carbon are in the simple ratio of } 2:1\text{.}$
3. $\text{When magnesium burns in oxygen, the amount of magnesium used for the reaction is equal to the amount of magnesium formed in magnesium oxide.}$
4. $\text{At constant temperature and pressure, } 200 \text{ mL of hydrogen will combine with } 100 \text{ mL of oxygen to produce } 200 \text{ mL of water vapour.}$

Solution:

$\text{Hint: Law of multiple proportion.}$ $\text{Explanation:}$ $\text{The element, carbon, combines with oxygen to form two compounds, namely, carbon dioxide and carbon monoxide. In } \text{CO}_2\text{, } 12 \text{ parts by mass of carbon combine with } 32 \text{ parts by mass of oxygen while in } \text{CO}\text{, } 12 \text{ parts by mass of carbon combine with } 16 \text{ parts by mass of oxygen.}$ $\text{Therefore, the masses of oxygen combine with a fixed mass of carbon (}12 \text{ parts) in } \text{CO}_2 \text{ and } \text{CO} \text{ are } 32 \text{ and } 16 \text{ respectively.}$ $\text{These masses of oxygen bear a simple ratio of } 32 : 16 \text{ or } 2 : 1 \text{ to each other. This is an example of law of multiple proportion.}$

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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
}