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Current Question (ID: 10560)

Question:

$\text{A brown-coloured mixture of two gases is obtained by reducing 6 N nitric acid with metallic copper. This mixture on cooling condenses to a blue liquid which on freezing (-30 °C) gives a blue solid. The correct statement for the blue liquid or solid is:}$

Options:

1. $\text{It is referred to as the anhydride of nitrous acid.}$
2. $\text{It is an acidic oxide. Hence, it dissolves in alkalies producing nitrites.}$
3. $\text{It can also be prepared by the action of 50 \% } \text{HNO}_3 \text{ on arsenious oxide and then cooling to 250 K.}$
4. $\text{All of the above.}$ (Correct)

Solution:

$\text{Hint:}$ $\text{The reactions are as follows:}$ $2\text{Cu} + 6\text{HNO}_3 \rightarrow 2\text{Cu}(\text{NO}_3)_2 + \text{NO} + \text{NO}_2 + 3\text{H}_2\text{O}$ $\text{The NO and } \text{NO}_2 \text{ mixture is known as blue liquid or solid. It is an anhydride of nitrous acid.}$ $\text{NO} + \text{NO}_2 \rightleftharpoons \text{N}_2\text{O}_3\text{(blue)} \text{ at } -20°\text{C}$ $(1) 2\text{HNO}_2 \rightarrow \text{N}_2\text{O}_3 + \text{H}_2\text{O}$ $(2) \text{N}_2\text{O}_3 + 2\text{KOH} \rightarrow 2\text{KNO}_2 + \text{H}_2\text{O}$ $\text{It can also be prepared by the action of 50\% } \text{HNO}_3 \text{ on arsenious oxide and then cooling to 250K. The reaction is as follows:}$ $2\text{HNO}_3 + \text{As}_2\text{O}_3 + 2\text{H}_2\text{O} \rightarrow \text{NO} + \text{NO}_2 + 2\text{H}_3\text{AsO}_4$ $\downarrow \text{ 250K}$ $\text{N}_2\text{O}_3$ $\text{All options are correct.}$

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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
}