JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of \(3.33\, h\) at \(25^{\circ} C\). After \(9 \,h ,\) the fraction of sucrose remaining is \(f\). The value of \(\log _{10}\left(\frac{1}{f}\right)\) is ..... \(\times 10^{-2} .\) (Rounded off to the nearest integer) [Assume : \(\ln 10=2.303, \ln 2=0.693\) ]
- A \(475\)
- B \(525\)
- C \(125\)
- D \(81\)
Answer & Solution
Correct Answer
(D) \(81\)
Step-by-step Solution
Detailed explanation
Given : \(C _{12} H _{22} O _{11}+ H _{2} O \frac{ I \text { order }}{ t _{1 / 2}=\frac{10}{3} hr } \underset{\text { Glucose }}{ C _{6} H _{12} O _{6}}+\underset{\text { Fructose }}{ C _{6} H _{12} O _{6}}\)…
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