Question

Sucrose decomposes in solution into glucose and fructose according to the first order rate law. With $t^{1/2}3$ Hr. What fraction of sample of sucrose remains after $8$ hour?

Answer 1

$t_{\frac{1}{2}}=3hrs$

$or,\frac{0.93}{K}=3$

$or,K=0.231$

let, $A_0$ be the initial concentration.

K$=\frac{2.303}{t}\log \left[\frac{A_0}{A_8}\right]$

$or,0.231=\frac{2.303}{t}\log \left[\frac{A_0}{A_8}\right]$

$or,{\log }_{10}\left[\frac{A_0}{A_8}\right]=0.802$

$or,\frac{A_0}{A_{8hours}}=6.33$

$\therefore$ concentration remaining after 8 hours$A_{8hours}=\frac{A_0}{6.33}$