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Current Question (ID: 20850)

Question:

$\text{When } 35 \text{ mL of } 0.15 \text{ M lead nitrate solution is mixed with } 20 \text{ mL of } 0.12 \text{ M chromic sulphate solution, } x \times 10^{-5} \text{ moles of lead sulfate precipitate out. Calculate the value of } x. \text{ (Round off to the Nearest Integer)}$

Options:

1. $525$
2. $124$
3. $734$
4. $223$

Solution:

$\text{Hint: The limiting reagent is } \text{Pb(NO}_3\text{)}_2$ $\text{Step 1: The balanced reaction is given as follows:}$ $3 \text{Pb(NO}_3\text{)}_2 + \text{Cr}_2\text{(SO}_4\text{)}_3 \rightarrow 3 \text{PbSO}_4 + 2 \text{Cr(NO}_3\text{)}_3$ $\text{The volume of lead nitrate solution } = 35 \text{ mL } = 0.035 \text{ L}$ $\text{The molarity of lead nitrate solution } = 0.15 \text{ M } = 0.15 \text{ mol L}^{-1}$ $\text{So, the number of moles of lead nitrate } = 0.15 \text{ mol L}^{-1} \times 0.035 \text{ L } = 0.00525 \text{ mol } = 5.25 \text{ millimoles}$ $\text{Step 2: The volume of chromic sulphate solution } = 20 \text{ mL } = 0.020 \text{ L}$ $\text{The molarity of chromic sulphate solution } = 0.12 \text{ M } = 0.12 \text{ mol L}^{-1}$ $\text{So, the number of moles of chromic sulphate } = 0.12 \text{ mol L}^{-1} \times 0.020 \text{ L } = 0.0024 \text{ mol } = 2.4 \text{ millimoles}$ $\text{Step 3: 3 moles of lead nitrate react with one mole of chromic sulphate.}$ $\text{1 mole of lead nitrate reacts with } \frac{1}{3} \text{ moles of chromic sulphate.}$ $\text{Hence, 5.25 moles of lead nitrate reacts with } \frac{5.25}{3} = 1.75 \text{ moles of chromic sulphate.}$ $\text{The limiting reagent is } \text{Pb(NO}_3\text{)}_2 \text{ and the excess reactant is } \text{Cr}_2\text{(SO}_4\text{)}_3$ $\text{The millimole of } \text{PbSO}_4 = 5.25 \text{ because the molar ratio of } \text{Pb(NO}_3\text{)}_2 \text{ and } \text{PbSO}_4 \text{ is } 1 : 1.$ $\text{mole of } \text{PbSO}_4 = 525 \times 10^{-5}$ $\text{So } x = 525$ $\text{Hence, option 1 is the correct answer.}$

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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
}