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Current Question (ID: 7433)

Question:

Concentrated nitric acid is 70\% $\text{HNO}_3$. The amount of concentrated nitric acid solution that should be used to prepare 250 mL of 2.0 M $\text{HNO}_3$ would be:

Options:

1. \text{90.0 g conc. HNO}_3
2. \text{70.0 g conc. HNO}_3
3. \text{54.0 g conc. HNO}_3
4. \text{45.0 g conc. HNO}_3

Solution:

\text{Hint: Molarity} = \frac{\text{weight of } \text{HNO}_3}{\text{Molecular mass of } \text{HNO}_3 \times \text{volume of solution(L)}}\n\n\text{Step 1:}\n\n\text{Given,}\n\n\text{molarity of solution} = 2.0 \text{ M}\n\n\text{Volume of solution} = 250 \text{ mL} = 250/1000 = 1/4 \text{ L}\n\n\text{Molar mass of } \text{HNO}_3 = 1+14+3 \times 16 = 63 \text{ g mol}^{-1}\n\n\text{Molarity} = \frac{\text{weight of } \text{HNO}_3}{\text{Molecular mass of } \text{HNO}_3 \times \text{volume of solution(L)}}\n\n\text{Weight of } \text{HNO}_3 = \text{molarity} \times \text{molecular mass} \times \text{volume(L)}\n\n= 2 \times 63 \times 1/4 \text{ g}\n\n= 31.5 \text{ g}\n\n\text{Step 2:}\n\n\text{It is the weight of 100\% } \text{HNO}_3\n\n\text{70\% } \text{HNO}_3 \text{ means 70 g } \text{HNO}_3 \text{ present in 100 gram solution.}\n\n\text{So, amount of concentrated (70\%) } \text{HNO}_3 \text{ solution required} = \frac{31.5 \times 100}{70} = 45.0 \text{ g}

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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
}