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Current Question (ID: 18128)

Question:

$\text{The amount of CaCl}_2 \ (i = 2.47) \ \text{dissolved in 2.5 litre of water such that its}$ $\text{osmotic pressure is 0.75 atm at 27 } ^\circ \text{C is:}$

Options:

1. $1.02 \ \text{g}$
2. $4.35 \ \text{g}$
3. $2.87 \ \text{g}$
4. $3.42 \ \text{g}$

Solution:

$\text{HINT: Use general expression of osmotic pressure}$ $\text{STEP 1: We know that,}$ $\pi = i \frac{w}{MV}RT$ $\Rightarrow \pi = \frac{i w}{MV}RT$ $\Rightarrow w = \frac{\pi MV}{iRT}$ $\pi = 0.75 \ \text{atm}$ $V = 2.5 \ \text{L}$ $i = 2.47$ $T = (27 + 273) \ \text{K} = 300 \ \text{K}$ $\text{Here, } R = 0.0821 \ \text{L atm K}^{-1}\text{mol}^{-1} ; M = 1 \times 40 + 2 \times 35.5 = 111 \text{g mol}^{-1}$ $\text{STEP 2:}$ $\text{Therefore, } w = \frac{0.75 \times 111 \times 2.5}{2.47 \times 0.0821 \times 300} = 3.42 \ \text{g}$ $\text{Hence, the required amount of CaCl}_2 \text{ is 3.42 g.}$

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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
}