Question

Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of $3.33 h$ at $25^{\circ} C$. After $9 h$, 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

Given :



from I order kinetic : $\frac{ k \times t }{2.303}=\log \frac{| A |_{0}}{| A |_{ t }}$

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

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

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

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

$\Rightarrow x =81$