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Current Question (ID: 20830)

Question:
$250 \text{ mL solution of } \text{CH}_3\text{COONa of molarity } 0.35 \text{ M is prepared.}$ $\text{What is the mass of } \text{CH}_3\text{COONa required in grams? (Nearest integer)}$
Options:
  • 1. $\text{Nine (9)}$
  • 2. $\text{Seven (7)}$
  • 3. $\text{Eight (8)}$
  • 4. $\text{Ten (10)}$
Solution:
$\text{Molarity} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in litre}}$ $\text{Moles of solute} = \frac{\text{Weight}}{\text{Molecular weight}}$ $0.35 = \frac{W}{\text{MW (CH}_3\text{COONa)}} \times \frac{1000}{25}$ $W = \frac{0.35 \times 82 \times 250}{1000}$ $W = \frac{7175}{1000} = 7.175 \text{ g}$ $\text{Mass of } \text{CH}_3\text{COONa required to prepare 250 mL of 0.35 M solution is 7.175 g i.e, approximately 7 g.}$

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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
}