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Question:
$\text{Suppose the elements X and Y combine to form two compounds } \text{XY}_2 \text{ and } \text{X}_3\text{Y}_2\text{. If 0.1 mole of } \text{XY}_2 \text{ weighs 10 g and 0.05 mole of } \text{X}_3\text{Y}_2 \text{ weighs 9 g, the atomic weight of X and Y respectively, will be:}$
Options:
  • 1. $40, 30$
  • 2. $60, 40$
  • 3. $20, 30$
  • 4. $30, 20$
Solution:
$\text{Hint: We can use the relationship between moles, mass and molar mass to set up equations and solve for the atomic weights.}$\n\n$\text{Step 1: Let's denote the atomic masses of X and Y as } A_X \text{ and } A_Y \text{ respectively.}$\n\n$\text{Step 2: For the compound } \text{XY}_2\text{, calculate the molar mass.}$\n\n$\text{Molar mass of } \text{XY}_2 = A_X + 2A_Y$\n\n$\text{Given that 0.1 mole of } \text{XY}_2 \text{ weighs 10 g:}$\n\n$\text{Mass} = \text{Number of moles} \times \text{Molar mass}$\n\n$10 \text{ g} = 0.1 \text{ mol} \times (A_X + 2A_Y)$\n\n$A_X + 2A_Y = \frac{10 \text{ g}}{0.1 \text{ mol}} = 100 \text{ g/mol} \ldots \text{(i)}$\n\n$\text{Step 3: For the compound } \text{X}_3\text{Y}_2\text{, calculate the molar mass.}$\n\n$\text{Molar mass of } \text{X}_3\text{Y}_2 = 3A_X + 2A_Y$\n\n$\text{Given that 0.05 mole of } \text{X}_3\text{Y}_2 \text{ weighs 9 g:}$\n\n$9 \text{ g} = 0.05 \text{ mol} \times (3A_X + 2A_Y)$\n\n$3A_X + 2A_Y = \frac{9 \text{ g}}{0.05 \text{ mol}} = 180 \text{ g/mol} \ldots \text{(ii)}$\n\n$\text{Step 4: Solve the system of equations (i) and (ii) to find } A_X \text{ and } A_Y\text{.}$\n\n$\text{Subtract equation (i) from equation (ii):}$\n\n$(3A_X + 2A_Y) - (A_X + 2A_Y) = 180 - 100$\n\n$2A_X = 80$\n\n$A_X = 40 \text{ g/mol}$\n\n$\text{Substitute } A_X = 40 \text{ into equation (i):}$\n\n$40 + 2A_Y = 100$\n\n$2A_Y = 60$\n\n$A_Y = 30 \text{ g/mol}$\n\n$\text{Therefore, the atomic weights of X and Y are 40 and 30 g/mol, respectively.}$

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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
}