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 nearest integer)

K=0.693t1/2⇒0.6933.33⇒0.208K=2.303tlogR0R0.208=2.3039log1flog1f=0.813=81.3×10−2

Book Online Demo

Check Your IQ

Try Test

Courses

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

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 81.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

$K = \frac{0.693}{t_{1 / 2}} \Rightarrow \frac{0.693}{3.33} \Rightarrow 0.208$

$K = \frac{2.303}{t} \log (\frac{R_{0}}{R})$

$0.208 = \frac{2.303}{9} \log (\frac{1}{f})$

$\log (\frac{1}{f}) = 0.813 = 81.3 \times 10^{- 2}$

Watch 3-min video & get full concept clarity