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Current Question (ID: 20786)

Question:

$1 \text{ gram of a carbonate } (\text{M}_2\text{CO}_3) \text{ on treatment with excess HCl produces } 0.01186 \text{ moles of } \text{CO}_2. \text{ The molar mass of } \text{M}_2\text{CO}_3 \text{ in g mol}^{-1} \text{ is:}$

Options:

1. $118.6$
2. $11.86$
3. $88.6$
4. $84.3$

Solution:

$\text{Hint: The formula of number of mole } = \frac{\text{amount of substance}}{\text{molar mass of substance}}$ $\text{Step 1:}$ $\text{The chemical reaction is as follows:}$ $\text{M}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{MCl} + \text{CO}_2 + \text{H}_2\text{O}$ $\text{The metal M shows +1 oxidation state because } \text{CO}_3^{2-} \text{ has a -2 charge and for balancing this charge two M combines with } \text{CO}_3^{2-}.$ $\text{From the reaction, it is clear that 1 mole of } \text{M}_2\text{CO}_3 \text{ gives 1 mol of } \text{CO}_2.$ $\text{Hence, } 0.01186 \text{ mol } \text{CO}_2 \text{ is generated from } 0.01186 \text{ mol } \text{M}_2\text{CO}_3.$ $\text{Step 2:}$ $\text{The formula of number of mole } = \frac{\text{amount of substance}}{\text{molar mass of substance}}$ $0.01186 = \frac{1}{\text{molar mass of } \text{M}_2\text{CO}_3}$ $\text{molar mass of } \text{M}_2\text{CO}_3 = \frac{1}{0.01186}$ $= 84.32 \text{ g mol}^{-1}$

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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
}