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Current Question (ID: 20779)

Question:

$\text{Complex A has a composition of } \text{H}_{12}\text{O}_6\text{Cl}_3\text{Cr} . \text{ If the complex on treatment with concentrated } \text{H}_2\text{SO}_4 \text{ loses 13.5\% of its original mass, the correct molecular formula of A is:}$ $\text{[Given: atomic mass of Cr = 52 amu and Cl = 35 amu]}$

Options:

1. $[\text{Cr(H}_2\text{O})_5\text{Cl}]\text{Cl}_2\cdot 2\text{H}_2\text{O}$
2. $[\text{Cr(H}_2\text{O})_6]\text{Cl}_3$
3. $[\text{Cr(H}_2\text{O})_3\text{Cl}_3]\cdot 3\text{H}_2\text{O}$
4. $[\text{Cr(H}_2\text{O})_4\text{Cl}_2]\text{Cl}\cdot 2\text{H}_2\text{O}$

Solution:

$\text{Hint: 13.5\% of its original mass is equal to 36 g}$ $\text{Step 1:}$ $\text{Calculate the molar mass of } \text{H}_{12}\text{O}_6\text{Cl}_3\text{Cr}.$ $\text{The molar mass of } \text{H}_{12}\text{O}_6\text{Cl}_3\text{Cr} = 12 \times (1) + 6 \times (16) + 3 \times (35) + 52$ $= 266.$ $\text{Calculate the 13.5\% of losses from its original mass is as follows:}$ $13.5 = \frac{x}{266} \times 100$ $\frac{13.5 \times 266}{100} = x$ $x = 36 \text{ g}$ $36 \text{ g of substance is lost.}$ $\text{Step 2:}$ $\text{The amount of 1 mole of water is 18 g and two moles of water is 36 g. Hence, it indicates that 2 water molecules are present outside the coordination sphere.}$ $\text{Hence, the correct formula of the compound is } [\text{Cr(H}_2\text{O})_4\text{Cl}_2]\text{Cl}\cdot 2\text{H}_2\text{O}.$

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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
}