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Current Question (ID: 18109)

Question:

$\text{The depression in the freezing point of water when 10 g of CH}_3\text{CH}_2\text{CHClCOOH is added to 250 g of water will be:}$ $\left( K_a = 1.4 \times 10^{-3}, K_f = 1.86 \text{ K kg mol}^{-1} \right)$

Options:

1. 0.32 \text{ K}
2. 2.87 \text{ K}
3. 0.65 \text{ K}
4. 5.03 \text{ K}

Solution:

$\text{HINT: Use } \Delta T_f = i \cdot K_f \cdot m$ $\text{STEP 1: Molar mass of CH}_3\text{CH}_2\text{CHClCOOH} = 122.5 \text{ g mol}^{-1}$ $\therefore \text{No. of moles present in 10 g of CH}_3\text{CH}_2\text{CHClCOOH} = \frac{10}{122.5} \text{ g mol}^{-1} = 0.0816 \text{ mol}$ $\text{It is given that 10 g is added to 250 g of water.}$ $\text{Molality of the solution, } m = \frac{0.0816}{250} \times 1000 = 0.3264$ $\text{STEP 2: Let } \alpha \text{ be the degree of dissociation of CH}_3\text{CH}_2\text{CHClCOOH.}$ $\text{CH}_3\text{CH}_2\text{CHClCOOH} \rightarrow \text{CH}_3\text{CH}_2\text{CHClCOO}^- + \text{H}^+$ $\text{Initial conc. } C \text{ mol L}^{-1} \quad 0 \quad 0$ $\text{At equilibrium } C(1-\alpha) \quad C\alpha \quad C\alpha$ $\therefore K_a = \frac{C\alpha \cdot C\alpha}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}$ $\text{Since } \alpha \text{ is very small with respect to 1, } 1-\alpha \approx 1$ $\text{Now, } K_a = C\alpha^2$ $\Rightarrow \alpha = \sqrt{\frac{K_a}{C}}$ $\alpha = \sqrt{\frac{1.4 \times 10^{-3}}{0.3264}} = 0.0655$ $\text{STEP 3: Again,}$ $\text{CH}_3\text{CH}_2\text{CHClCOOH} \rightarrow \text{CH}_3\text{CH}_2\text{CHClCOO}^- + \text{H}^+$ $\text{Initial conc. } 1 \quad 0 \quad 0$ $\text{At equilibrium } 1-\alpha \quad \alpha \quad \alpha$ $\text{Total moles of equilibrium } = 1-\alpha+\alpha+\alpha = 1+\alpha$ $\therefore i = \frac{1+\alpha}{1} = 1+\alpha = 1+0.0655 = 1.0655$ $\text{STEP 4: Hence, the depression in the freezing point of water is given as:}$ $\Delta T_f = i \cdot K_f \cdot m$ $= 1.0655 \times 1.86 \text{ K kg mol}^{-1} \times 0.3264 \text{ mol kg}^{-1}$ $= 0.65 \text{ K}$

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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
}