Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25°C . After 9h, the fraction of sucrose remaining is f . The value of log101f is ____________×10−2. (Rounded off to the nearest integer) Assume:ln 10=2.303, ln 2=0.693

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answer is 82.

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Detailed Solution

C12H22O11+H2O→C2H12O6+C6H12O6t12=3.33 hours 93.33=2.7Fraction of sucrose remains 122.7log1f=log22.7 =2.7×0.3010=0.82 =0.82×100 = 82

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