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Current Question (ID: 20785)

Question:

$5 \text{ moles of } \text{AB}_2 \text{ weight } 125 \times 10^{-3} \text{ kg and } 10 \text{ moles } \text{A}_2\text{B}_2 \text{ weight } 300 \times 10^{-3} \text{ kg. The molar mass of A } (M_A) \text{ and molar mass of B } (M_B) \text{ in kg mol are:}$

Options:

1. $M_A = 10 \times 10^{-3} \text{ and } M_B = 5 \times 10^{-3}$
2. $M_A = 25 \times 10^{-3} \text{ and } M_B = 50 \times 10^{-3}$
3. $M_A = 5 \times 10^{-3} \text{ and } M_B = 10 \times 10^{-3}$
4. $M_A = 50 \times 10^{-3} \text{ and } M_B = 25 \times 10^{-3}$

Solution:

$\text{Hint: Number of mole} = \frac{\text{amount of substance}}{\text{molar mass of substance}}$ $\text{Step 1:}$ $\text{Calculate the molar mass of } \text{AB}_2 \text{ as follows:}$ $\text{molar mass of } \text{AB}_2 (g/mol) = \frac{125}{5} = 25$ $M_A + 2M_B = 25 \ldots \ldots \ldots (1)$ $10 = \frac{300}{2M_A + 2M_B}$ $2M_A + 2M_B = 30 \ldots \ldots \ldots (2)$ $\text{Step 2:}$ $\text{Solving equations 1 and 2 we get,}$ $M_A = 5 \text{ g/mol} = 5 \times 10^{-3} \text{ kg/mol}$ $M_B = 10 \text{ g/mol} = 10 \times 10^{-3} \text{ kg/mol}$

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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
}