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Current Question (ID: 7425)

Question:

$25.3 \text{ g of sodium carbonate, } \text{Na}_2\text{CO}_3\text{, is dissolved in enough water to make } 250 \text{ mL of a solution. If sodium carbonate completely dissociates, the molar concentrations of sodium ion, } \text{Na}^+ \text{ and carbonate ion, } CO_3^{2-} \text{ are respectively:}$ $(\text{molar mass of } \text{Na}_2\text{CO}_3 = 106 \text{ g mol}^{-1})$

Options:

1. $0.955 \text{ M and } 1.910 \text{ M}$
2. $1.910 \text{ M and } 0.955 \text{ M}$
3. $1.90 \text{ M and } 1.910 \text{ M}$
4. $0.477 \text{ M and } 0.477 \text{ M}$

Solution:

$\text{Molarity} = \frac{\text{number of moles of solute}}{\text{volume of solution (in mL)} \times 1000}$ $= \frac{25.3 \times 1000}{106 \times 250} = 0.9547 \approx 0.955 \text{ M}$ $\text{Na}_2\text{CO}_3 \text{ in aqueous solution remains dissociated as}$ $\text{Na}_2\text{CO}_3 \rightarrow 2\text{Na}^+ + \text{CO}_3^{2-}$ $\quad x \quad \quad 2x \quad \quad x$ $\text{Since, the molarity of } \text{Na}_2\text{CO}_3 \text{ is } 0.955 \text{ M, the molarity of } \text{CO}_3^{2-} \text{ is also } 0.955 \text{ M and that of } \text{Na}^+ \text{ is } 2 \times 0.955 = 1.910 \text{ M}$

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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
}