Sucrose Decomposes In Acid Solution Into Glucose And Fructose According To The First Order Rate Law, With T1/2 = 3.00 Hours. What Fraction Of Sample Of Sucrose Remains After 8 Hours?

\( k = \frac{0.063}{t_{1/2}} \) \( \Rightarrow \frac{0.063}{3.0} \) \( \Rightarrow 0.231/hr \)

Let initial concetration of sucrose be 1 M.

After 8 hours, the concentration will be 1−x. X represents the amount of sucrose decomposed.

\( k = \frac{2.303}{t} log\frac{[A]_{0}}{[A]} \) \( \Rightarrow 0.231 = \frac{2.303}{8} log\frac{[1]}{[1- x]} \) \( \Rightarrow log\frac{[1]}{[1- x]} = 0.8024 \) \( \Rightarrow \frac{1}{1- x} = 6.345 \) \( \Rightarrow x = 0.842 \)

The concentration of a sample of sucrose after 8 hours will be

\( \Rightarrow 1 – 0.842 = 0.158 M. \)

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