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JEE MAIN Chemistry QUESTION #5534
Question 1
Sucrose undergoes acid hydrolysis via first-order kinetics with a half-life of 3.33 h at 25°C. After 9 h, the fraction of sucrose remaining is $f$. Find the value of $\log_{10}\!\left(\dfrac{1}{f}\right)\times10^{-2}$ (nearest integer). [Given: $\ln 10 = 2.303$, $\ln 2 = 0.693$]
  • 81✔️
  • 90
  • 135
  • 27
Correct Answer Explanation
$k = \frac{0.693}{3.33} = 0.208\ h^{-1}$. After $t=9$ h: $\ln\left(\frac{1}{f}\right) = kt = 0.208\times9 = 1.872$. $\log_{10}\left(\frac{1}{f}\right) = \frac{1.872}{2.303} = 0.8127$. Answer $= 0.8127\times10^{-2}\times10^{2}\approx81\times10^{-2}$, so the value is $\approx 81\times10^{-2}$, meaning the answer = 81.