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Current Question (ID: 21508)

Question:

$\text{Sucrose hydrolysis in an acidic solution into glucose and fructose follows the first-order rate law with a half-life of 3.33 h at 25}^\circ\text{C. After 9 h, the fraction of sucrose remaining is } f. \text{ The value of } \log\left(\frac{1}{f}\right) \text{ is } A \times 10^{-2}. \text{ The value of } A \text{ is: (Rounded off to the nearest integer) [Assume : ln 10 = 2.303, ln 2 = 0.693]}$

Options:

1. $78$
2. $81$
3. $85$
4. $75$

Solution:

$\text{Hint: } k = \frac{2.303}{t} \log\frac{1}{f}$ $\text{Step 1:}$ $\text{The reaction is as follows:}$ $\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \xrightarrow{\text{I order } t_{1/2} = \frac{10}{3} \text{ hr}} \text{C}_6\text{H}_{12}\text{O}_6 \text{ (Glucose) } + \text{C}_6\text{H}_{12}\text{O}_6 \text{ (Fructose)}$ $t = 0 \quad a = [A]_0$ $t = 9 \text{ hr} \quad a - x = [A]_t$ $\text{Calculate the value of } k \text{ using } t_{1/2} \text{ value.}$ $t_{1/2} = \frac{0.693}{k}$ $k = \frac{0.693}{3.33} = 0.208 \text{ h}^{-1}$ $\text{Step 2:}$ $\text{Calculate the value of } \frac{1}{f} \text{ using a first-order equation as follows:}$ $\frac{k \times t}{2.303} = \log\frac{[A]_0}{[A]_t}$ $\text{The initial concentration of sucrose is 1. After 9 h the concentration of sucrose is } [A]_t$ $\Rightarrow 0.6 = \frac{93 \times 9}{2.303} = \log\left(\frac{1}{f}\right)$ $\Rightarrow \frac{0.693 \times 9 \times 3}{23.03} = \log\left(\frac{1}{f}\right)$ $\Rightarrow \log\left(\frac{1}{f}\right) = 0.81246 = 81.24 \times 10^{-2}$ $\Rightarrow x = 81$

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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
}