Import Question JSON

Current Question (ID: 8477)

Question:

$\text{The solution with pH value close to 1.0 among the following is:}$

Options:

1. $100 \text{ ml of } (\text{M}/10) \text{HCl} + 100 \text{ ml of } (\text{M}/10) \text{NaOH}$
2. $55 \text{ ml of } (\text{M}/10) \text{HCl} + 45 \text{ ml of } (\text{M}/10) \text{NaOH}$
3. $10 \text{ ml of } (\text{M}/10) \text{HCl} + 90 \text{ ml of } (\text{M}/10) \text{NaOH}$
4. $85 \text{ ml of } (\text{M}/10) \text{HCl} + 15 \text{ ml of } (\text{M}/10) \text{NaOH}$

Solution:

$\text{Hint: } [\text{H}^+] = \frac{\text{M}_1\text{V}_1 - \text{M}_2\text{V}_2}{\text{V}_1 + \text{V}_2}$ $\text{Step 1:}$ $\text{In the first option, the volume and concentration of acid and base are the same, hence, it is a neutral solution.}$ $\text{In the second option, the volume of acid is more than base, hence, it is an acidic solution.}$ $\text{In the third option, the volume of the base is more than acid, hence, it is a basic solution.}$ $\text{Step 2:}$ $\text{Calculate the pH for 4}^{\text{th}} \text{ option as follows:}$ $[\text{H}^+] = \frac{\text{M}_1\text{V}_1 - \text{M}_2\text{V}_2}{\text{V}_1 + \text{V}_2}$ $= \frac{0.1 \times 85 - 0.1 \times 15}{100} = \frac{8.5 - 1.5}{100}$ $[\text{H}^+] = 7/100 = 7 \times 10^{-2} \text{ M}$ $\text{pH} = -\log[\text{H}^+] = -\log(7 \times 10^{-2})$ $= 2 - \log 7$ $= 2 - 0.85$ $= 1.15$ $\text{i.e., in this case, pH is close to 1.}$

Import JSON File

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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
}