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Current Question (ID: 21401)

Question:

$\text{If } 250 \text{ cm}^3 \text{ of an aqueous solution containing } 0.73 \text{ g of a protein A is isotonic with one litre of another aqueous solution containing } 1.66 \text{ g of a protein B at } 298 \text{ K, the ratio of the molecular masses of A and B is:}$

Options:

1. $1.54$
2. $1.77$
3. $1.65$
4. $1.60$

Solution:

$\text{Let the molar mass of protein A = } x \text{ g/mol}$ $\text{Let the molar mass of protein B = } y \text{ g/mol}$ $\pi_A = \text{ osmotic pressure of protein A } = \frac{0.73}{x \times 0.25} RT$ $\pi_B = \text{ osmotic pressure of protein B } = \frac{1.65}{y} RT$ $\text{Isotonic solution has same osmotic pressure}$ $\pi_A = \pi_B$ $\left( \frac{x}{y} \right) = \frac{0.73}{0.25 \times 1.65}$ $= 1.769 = 1.77$

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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
}