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Current Question (ID: 18708)

Question:

$\text{An excess of AgNO}_3 \text{ is added to } 100 \text{ mL of a } 0.01\text{M solution of dichlorotetraaquachromium(III) chloride.}$ $\text{The number of moles of AgCl precipitated would be:}$

Options:

1. $0.002$
2. $0.003$
3. $0.01$
4. $0.001$

Solution:

$\text{HINT: One mole of Cl}^- \text{ will react with AgNO}_3 \text{ to give one mole of AgCl.}$ $\text{STEP 1: The given complex is } [\text{Cr(H}_2\text{O)}_4\text{Cl}_2]^+\text{Cl}^-, \text{hence only one ionizable Cl}^- \text{ is present and will be precipitated.}$ $\text{The reaction will be: } \left[ \text{Cr(H}_2\text{O)}_4\text{Cl}_2 \right]\text{Cl} + \text{AgNO}_3 \rightarrow \left[ \text{Cr(H}_2\text{O)}_4\text{Cl}_2 \right]^+ + \text{AgCl} + \text{NO}_3^- $ $\text{STEP 2: } 0.01 \text{ M means } 0.01 \text{ mol per } 1000 \text{ ml of solution.}$ $100 \text{ ml(given) sol will have } \frac{0.01}{1000} \times 100 = 0.001 \text{ mol of complex}$ $\text{Hence, } 0.001 \text{ mol of AgCl will be precipitated.}$

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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
}