Question

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

Answer 1

Answer: 81

Given :



from I order kinetic : $\frac{k\times t}{2.303}=\log \frac{|A{|}_0}{|A{|}_t}$

$\Rightarrow \frac{\ln 2\times 9}{\frac{10}{3}\times 2.303}=\log \left(\frac{1}{f}\right)$

$\Rightarrow \frac{0.693\times 9\times 3}{23.03}=\log \left(\frac{1}{f}\right)$

$\Rightarrow \log \left(\frac{1}{f}\right)=0.81246$

$=81.24\times 10^{-2}$

$\Rightarrow x=81$