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Current Question (ID: 21412)

Question:

$\text{Consider separate solutions of } 0.500 \text{ M } \text{C}_2\text{H}_5\text{OH(aq)}, 0.100 \text{ M } \text{Mg}_3(\text{PO}_4)_2\text{(aq)}, 0.250 \text{ M } \text{KBr(aq)} \text{ and } 0.125 \text{ M } \text{Na}_3\text{PO}_4\text{(aq)} \text{ at } 25^\circ\text{C.}$ $\text{The correct statement about these solutions, assuming all salts to be strong electrolytes is:}$

Options:

1. $0.100 \text{ M } \text{Mg}_3(\text{PO}_4)_2\text{(aq)} \text{ has the highest osmotic pressure}$
2. $0.125 \text{ M } \text{Na}_3\text{PO}_4\text{(aq)} \text{ has the highest osmotic pressure}$
3. $0.500 \text{ M } \text{C}_2\text{H}_5\text{OH(aq)} \text{ has the highest osmotic pressure}$
4. $\text{They all have the same osmotic pressure}$

Solution:

$\text{Hint: The formula of osmotic pressure is } \pi = iCRT$ $\text{For all solution RT value is constant thus, osmotic pressure value}$ $\text{depends on } iC \text{ value. The } \alpha = 100\% \text{ for all the substance.}$ $0.500\text{M } \text{C}_2\text{H}_5 \text{OH (aq.)} \;;\; i = 1$ $0.100\text{M } \text{Mg}_3(\text{PO}_4)_2 \text{ (aq.)} \;;\; i = 5$ $0.250\text{M } \text{KBr(aq.)} \;;\; i = 2$ $0.125\text{M } \text{Na}_3 \text{PO}_4 \text{ (aq.)} \;;\; i = 4$ $\text{according to the formula of osmotic pressure,}$ $\pi \propto (i \times C)$ $\pi_{(\text{C}_2\text{H}_5 \text{OH})} \propto 1 \times 0.5$ $\pi_{(\text{Mg}_3(\text{PO}_4)_2)} \propto 5 \times 0.1$ $\pi_{(\text{KBr})} \propto 2 \times 0.25$ $\pi_{(\text{Na}_3 \text{PO}_4)} \propto 4 \times 0.125$ $\text{For all } \pi \text{ value is } 0.5RT. \text{ Hence, all have same osmotic pressure}$

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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
}