Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with $t_{1 / 2}=3$ hour. The percentage of sucrose remaining after 6 hours is

$\_\_\_\_$ . (Nearest integer)

(Given $: \log 2=0.3010$ and $\log 3=0.4771$ )

For a reaction $\mathrm{A} \rightarrow \mathrm{P}$ at T K , the half life $\left(\mathrm{t}_{1 / 2}\right)$ is plotted as a function of initial concentration $[\mathrm{A}]_0$ of A as given below.

The value of $x$ in the given figure is $\_\_\_\_$ s (Nearest integer)

If the half life of a first order reaction is 6.93 minutes then the time required for completion of $99 \%$ of the reaction will be $\_\_\_\_$ minutes.

(Given $: \log 2=0.3010$ )

$$ \text { For a first order reaction } \mathrm{A} \rightarrow \mathrm{~B} $$

$$ \begin{array}{|l|l|} \hline \mathrm{t} / \min & {[\mathrm{A}] / \mathrm{M}} \\ \hline 0 & 0.6500 \\ \hline x & 0.0650 \\ \hline 20 & 0.00065 \\ \hline \end{array} $$

$x=$ $\_\_\_\_$ min. (Nearest integer)