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Current Question (ID: 20797)

Question:

$\text{To neutralize 20 mL of 0.1 M aqueous solution of phosphorus acid } (\text{H}_3\text{PO}_3) \text{ completely, the volume of 0.1 M aqueous KOH solution required is:}$

Options:

1. $10 \text{ mL}$
2. $20 \text{ mL}$
3. $40 \text{ mL}$
4. $60 \text{ mL}$

Solution:

$\text{Hint: Use formula, } M_1V_1 = M_2V_2$ $\text{Step 1:}$ $\text{The reaction of } \text{H}_3\text{PO}_3 \text{ and NaOH is as follows:}$ $\text{H}_3\text{PO}_3 + 2\text{KOH} \rightarrow \text{K}_2\text{HPO}_3 + 2\text{H}_2\text{O}$ $\text{Calculate the initial mole of } \text{H}_3\text{PO}_3 \text{ as follows:}$ $\text{number of mole} = \text{Molarity} \times \text{volume of solution(L)}$ $= 0.1 \times 20 \times \frac{1}{1000}$ $= 0.002 \text{ mol}$ $\text{Step 2:}$ $\text{From the reaction, it is clear that 1 mole of } \text{H}_3\text{PO}_3 \text{ reacts with 2 moles of KOH. Hence, for 0.002 mol of } \text{H}_3\text{PO}_3 , 0.004 \text{ mol of KOH is used.}$ $\text{Calculate the volume of KOH solution is as follows:}$ $\text{Volume of KOH solution} = \frac{\text{number of moles}}{\text{Molarity}}$ $= \frac{0.004}{0.1}$ $= 0.04 \text{ L}$ $= 40 \text{ mL}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}