Sucrose hydrolyses in acidic medium into gluc | vedclass

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(A) For a first-order reaction, the amount remaining after $n$ half-lives is given by the formula $N = N_0 	imes (1/2)^n$.Here, the total time is $t

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) For a first-order reaction, the amount remaining after $n$ half-lives is given by the formula $N = N_0 	imes (1/2)^n$.Here, the total time is $t = 6 	ext{ hours}$ and the half-life is $t_{1/2} = 3 	ext{ hours}$.The number of half-lives $n$ is calculated as $n = t / t_{1/2} = 6 / 3 = 2$.The fraction of sucrose remaining is $(1/2)^n = (1/2)^2 = 1/4 = 0.25$.To find the percentage remaining, we multiply the fraction by $100$: $0.25 	imes 100 = 25\%$.Thus, $25\%$ of the sucrose remains after $6 	ext{ hours}$.

$S.No.$	$Time/s$	$Total Pressure/(atm)$
$1.$	$0$	$0.1$
$2.$	$115$	$0.28$

Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with $t_{1/2} = 3 \text{ hours}$. The percentage of sucrose remaining after $6 \text{ hours}$ is . . . . . . . (Nearest integer) (Given: $\log 2 = 0.3010$ and $\log 3 = 0.4771$)

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