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Current Question (ID: 21434)

Question:
$3 \text{ g of acetic acid is dissolved in } 500 \text{ g of water. The depression in the freezing point of the solution is given as } x \times 10^{-1} \text{ K. Calculate the value of } x \text{ to the nearest integer.}$ $\text{[ Given : } K_a \text{ of } \text{CH}_3\text{COOH} = 1.8 \times 10^{-5} \text{ and } K_f \text{ of water } = 1.86 \text{ K/molal}$ $\text{Density of water } = 1 \text{ g/mL ]}$
Options:
  • 1. $4$
  • 2. $6$
  • 3. $2$
  • 4. $1$
Solution:
$\text{Hint: } \Delta T_f = i \cdot K_f \cdot m$ $\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \text{ (Assuming } \alpha << 1 \text{ )}$ $\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.8 \times 10^{-5}}{10^{-1}}} = \sqrt{1.8 \times 10^{-4}}$ $= 1.3 \times 10^{-2} = 0.013$ $\text{So, } i = 1 + (2 - 1)(0.013)$ $= 1.013$ $\Delta T_f = 1.013 \times 1.86 \times \frac{3 \times 1000}{60 \times 500}$ $= 0.188$ $= 1.88 \times 10^{-1}$ $x \simeq 2$

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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
}