Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, ${\mathrm{t}}_{1/2}=3.00$ hours. What fraction of sample of sucrose remains after 8 hours?

${\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}+{\mathrm{H}}_2\mathrm{O}\stackrel{{\mathrm{H}}^{+}}{\to }{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6+{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$

It was $1^{\mathrm{st}}$ order reaction

${\mathrm{t}}_{1/2}=3 \mathrm{hrs}$

for$1^{\mathrm{st}} \mathrm{order},$ ${\mathrm{t}}_{1/2}=\frac{0.693}{\mathrm{k}}$

$\mathrm{k}=\frac{0.693}{{\mathrm{t}}_{1/2}}=\frac{0.693}{3}=0.23 1{\mathrm{hr}}^{-1}$

$\mathrm{K}=\frac{2.303}{\mathrm{t}}\log \frac{{\left[\mathrm{A}\right]}_0}{\left[\mathrm{A}\right]} \mathrm{say} {\left[\mathrm{A}\right]}_0=1$

$\left[\mathrm{A}\right]=1-x$

$0.231=\frac{2.303}{8 \mathrm{hr}}\log \left[\frac{1}{1-\mathrm{x}}\right]$

$\log \left[\frac{1}{1-\mathrm{x}}\right]=\frac{8\times 0.231}{2.303}=0.8024$

$\frac{1}{1-\mathrm{X}}={10}^{0.8024}=6.3445$

$1-\mathrm{X}=\frac{1}{6.2445}=0.158\mathrm{M}$