Ncert Exercises & Solutions

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours ?

For a first order reaction,

l = \frac{2.303}{t} \log \frac{{[R]}_{0}}{[R]}

It is given that,

t_{1 / 2}

= 3.00 hours

k = \frac{0.693}{t_{\frac{1}{2}}}

Therefore,

= 0.231 h^{- 1}

= \frac{2.303}{8 h} \log \frac{{[R]}_{0}}{[R]}

T h e n, 0.231 h^{- 1}

\Rightarrow \log \frac{{[R]}_{0}}{R} = \frac{0.231 h^{- 1} \times 8 h}{2.303}

\Rightarrow \frac{{[R]}_{0}}{R} = a n t i \log (0.8024)

\Rightarrow \frac{{[R]}_{0}}{R} = 6.3445

\Rightarrow \frac{{[R]}_{0}}{R} = 0.1576 (a p p r o x)

= 0.158

Hence, the fraction of the sample of sucrose that remains after 8 hours is 0.158.

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