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 $9 h,$ the fraction of sucrose remaining is $f .$ The value of $\log_{10} (\frac{1}{f})$ is ________ $\times 10^{- 2} .$ (Rounded off to the nearest integer)
[Assume: In $10 = 2.303, \ln 2 = 0.693$ ]

Question

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 $9 h,$ the fraction of sucrose remaining is $f .$ The value of $\log_{10} (\frac{1}{f})$ is ________ $\times 10^{- 2} .$ (Rounded off to the nearest integer)

[Assume: In $10 = 2.303, \ln 2 = 0.693$ ]

Aicharya · Accepted Answer

C 12 H 22 O 11 + H 2 O &#8594; t 1 / 2 = 10 3 hr order C 6 H 12 O 6 Glucose + C 6 H 12 O 6 Fructose t = 0 &#160;&#160; a = A 0 t = 9 hr &#160;&#160; a - x = A t from I order kinetic: k &#215; t 2 . 303 = log A 0 A t &#8658; ln 2 &#215; 9 10 3 &#215; 2 . 303 = log 1 f &#8658; 0 . 693 &#215; 9 &#215; 3 23 . 03 = log 1 f &#8658; log 1 f = 0 . 81246 = 81 . 24 &#215; 10 - 2 &#8658; x = 81

Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a…

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