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Current Question (ID: 7418)

Question:

$\text{When 50 mL of a 16.9\% (w/v) solution of AgNO}_3 \text{ is mixed with 50 mL of 5.8\% (w/v) NaCl solution, then the mass of precipitate formed is:}$ $(\text{Ag} = 107.8, \text{ N} = 14, \text{ O} = 16, \text{ Na}= 23, \text{ Cl}=35.5)$

Options:

1. $28 \text{ g}$
2. $3.5 \text{ g}$
3. $7 \text{ g}$
4. $14 \text{ g}$

Solution:

$\text{Hint: For the calculation of the mass of AgCl precipitated, we find mass of AgNO}_3 \text{ and NaCl in equal volume}$ $\text{with the help of the mole concept.}$ $16.9\% \text{ solution of AgNO}_3 \text{ means 16.9 g AgNO}_3 \text{ is present in 100 mL solution.}$ $8.45 \text{ g AgNO}_3 \text{ will present in 50 mL solution}$ $\text{Similarly,}$ $5.8 \text{ g NaCl is present in 100 mL solution}$ $2.9 \text{ g NaCl is present in 50 mL solution}$ $\text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3$ $\text{Initial mole} \quad 8.45/169.8 \quad 2.9/58.5 \quad 0 \quad 0$ $\quad = 0.049 \quad = 0.049$ $\text{After reaction} \quad 0 \quad 0 \quad 0.049 \quad 0.049$ $\text{Mass of AgCl precipitated} = 0.049 \times 143.5 = 7 \text{ g}$

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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
}