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Current Question (ID: 18111)

Question:

$\text{Two elements A and B form compounds having formula AB}_2 \text{ and AB}_4. \text{ When dissolved in 20 g of benzene (C}_6\text{H}_6), 1 \text{ g of AB}_2 \text{ lowers the freezing point by 2.3 K whereas 1.0 g of AB}_4 \text{ lowers it by 1.3 K. The atomic masses of A and B are respectively:}$ $\left(K_f \text{ for benzene is 5.1 K kg mol}^{-1}\right)$

Options:

1. $15.59 \text{ u and } 52.64 \text{ u}$
2. $25.59 \text{ u and } 42.64 \text{ u}$
3. $13.59 \text{ u and } 52.64 \text{ u}$
4. $23.59 \text{ u and } 32.64 \text{ u}$

Solution:

$\text{HINT: Use expression for depression in freezing point for both compounds.}$ $\text{STEP 1: We know that,}$ $M_2 = \frac{1000 \times w_2 \times k_f}{\Delta T_f \times w_1}$ $\text{Then, } M_{AB_2} = \frac{1000 \times 1 \times 5.1}{2.3 \times 20} = 110.87 \text{ g mol}^{-1}$ $M_{AB_4} = \frac{1000 \times 1 \times 5.1}{1.3 \times 20} = 196.15 \text{ g mol}^{-1}$ $\text{Now, we have the molar masses of AB}_2 \text{ and AB}_4 \text{ as 110.87 g mol}^{-1} \text{ and 196.15 g mol}^{-1} \text{ respectively.}$ $\text{STEP 2: Let the atomic masses of A and B be x and y respectively.}$ $\text{Now, we can write:}$ $x + 2y = 110.87 \quad (i)$ $x + 4y = 196.15 \quad (ii)$ $\text{Subtracting equation (i) from (ii), we have}$ $2y = 85.28$ $\Rightarrow y = 42.64$ $\text{Putting the value of 'y' in equation (1), we have}$ $x + 2 \times 42.64 = 110.87$ $\Rightarrow x = 25.59$ $\text{Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.}$

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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
}