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Current Question (ID: 18142)

Question:

$17.4 \% \text{ (w/v) } \text{K}_2\text{SO}_4 \text{ solution at } 27^\circ \text{C is isotonic to } 5.85 \% \text{ (w/v) NaCl solution at } 27^\circ \text{C.}$ $\text{If NaCl is 100 \% ionised, the \% ionisation of } \text{K}_2\text{SO}_4 \text{ in aqueous solution is-}$ $[\text{At wt. of K = 39, Na = 23}]$

Options:

1. $25 \%$
2. $75 \%$
3. $50 \%$
4. $\text{None of the above}$

Solution:

$\text{Hint: Isotonic solutions: A solutions that have the same osmotic pressure}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{Find the osmotic pressure of } \text{K}_2\text{SO}_4$ $\text{The formula of osmotic pressure is } \pi = iCRT$ $\text{K}_2\text{SO}_4 \Rightarrow 2\text{K}^+ + \text{SO}_4^{2-}$ $\text{for } 1 \quad 0 \quad 0$ $1-\alpha \quad 2\alpha \quad \alpha$ $i = 1 + 2\alpha$ $\therefore \pi_1 = \frac{\omega}{\text{V}} \text{ST} (1 + 2\alpha)$ $\pi_1 = \frac{17.4 \times 1000}{174 \times 100} \times \text{ST} (1 + 2\alpha)$ $= \text{ST} \times (1 + 2\alpha)$ $\text{Step 2}$ $\text{Find the osmotic pressure for NaCl}$ $\text{NaCl} \Rightarrow \text{Na}^+ + \text{Cl}^-$ $\text{for } 1 \quad 0 \quad 0$ $(1 - \alpha_1) \quad \alpha_1 \quad \alpha_1$ $\therefore \pi_2 = \frac{5.85 \times 1000}{58.5 \times 100} \times \text{ST} \times (1 + \alpha_1)$ $\therefore \alpha_1 = 1$ $\therefore \pi_2 = \text{ST} \times 2$ $\text{Step 3:}$ $\text{Given } \pi_1 = \pi_2 \text{ (isotonic solution)}$ $\therefore \text{ST} \times 2 = \text{ST} \times (1 + 2\alpha)$ $\text{or } \alpha = 0.5$ $\text{or 50\% ionisation of } \text{K}_2\text{SO}_4$

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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
}