91Ó°ÊÓ

Reacting with water in an acidic solution at 35°C, sucrose decomposes into glucose and fructose according to the law of uninhibited decay. An initial amount of 0.40 mole of sucrose decomposes to 0.36 mole in 30 minutes.

The amount of material present at a time t is given by the function $A\left(t\right)=A_0{e}^{kt}$where $k<0$ represents the decay rate.

Substitute localid="1647401702638" $A\left(t\right)=0.36,A_0=0.40,t=0.5$in the function $A\left(t\right)=A_0{e}^{kt}$

$$\begin{gathered} A\left(t\right)=A_0{e}^{kt} \\ 0.36=0.40e^{k×0.5} \\ \frac{0.36}{0.40}=e^{k×0.5} \\ \ln \left(0.9\right)=\ln \left(e^{k×0.5}\right) \\ \ln \left(0.9\right)=k×0.5 \\ k=\frac{\ln \left(0.9\right)}{0.5} \\ k=-0.211 \end{gathered}$$

The amount of sucrose remaining after 2 hours can be calculated by substituting

role="math" localid="1647401837394" $A_0=0.40,t=2,k=-0.211$in the function role="math" localid="1647401905449" $A\left(t\right)=A_0{e}^{kt}$

$$\begin{gathered} A\left(t\right)=A_0{e}^{kt} \\ A\left(t\right)=0.40e^{-0.211×2} \\ A\left(t\right)=0.26 \end{gathered}$$

Substitute $A_0=0.40,A\left(t\right)=0.10,k=-0.211$in the function $A\left(t\right)=A_0{e}^{kt}$

$$\begin{gathered} A\left(t\right)=A_0{e}^{kt} \\ 0.10=0.40e^{-0.211t} \\ \frac{0.10}{0.40}=e^{-0.211t} \\ \ln \left(0.25\right)=\ln \left(e^{-0.211t}\right) \\ \ln \left(0.25\right)=-0.211t \\ t=\frac{\ln \left(0.25\right)}{-0.211} \\ t=6.58hours \end{gathered}$$