QJEEM21Ch84e7: JEE Main 2021 Chemistry - Class 12 JEEM21Ch84e7 Solution: Sucrose hydrolyses in acid solution into glucose and fructose…

Question

Verified Solution • 2025–26 Session moderate

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( {{1 \over f}} \right)

is ________

\times

10^{$-$ 2}. (Rounded off to the nearest integer)

[Assume : ln 10 = 2.303, ln 2 = 0.693]

Accepted Answer

Given,

\underset{S u c r o s e}{C_{12} H_{22} O_{11}} + H_{2} O \overset{1 s t o r d e r}{⟶} \underset{G l u \cos e}{C_{6} H_{12} O_{6}} + \underset{F r u c t o s e}{C_{6} H_{12} O_{6}}

{t_{1/2}} = {{10} \over 3}h

At t = 0, a = [A]₀ (initial conc.)

t = 9h, a

-

x = [A]_t [conc. at time t]

For using 1st order equation,

K = {{2.303} \over t}\log {{{{[A]}_0}} \over {{{[A]}_t}}}

\Rightarrow {{K \times t} \over {2.303}} = \log {{{{[A]}_0}} \over {{{[A]}_t}}}

{{\ln 2 \times 9} \over {10/3 \times 2.303}} = \log \left( {{1 \over F}} \right) \Rightarrow \log \left( {{1 \over F}} \right) = 0.8124

(

\because

k = {{\ln 2} \over {{t_{1/2}}}}

)

\log \left( {{1 \over F}} \right) = 81.24 \times {10^{ - 2}}

x = 81.24 or x

\approx

81

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