The following data were obtained in experiment on inversion of cane sugar.
Time (minutes) 0 60 120 180 360 $\infty$
Angle of rotation +13.1 +11.6 +10.2 +9.0 +5.87 -3.8
(degree)
Determine total time ?

The correct option is C 966 min
The integrated rate law expression for the inversion of can sugar (assuming first order kinetics) is as shown.
$k=\frac{2.303}{t}log\frac{r_0-r_{\infty }}{r_t-r_{\infty }}$
For 60 minutes
$k=\frac{2.303}{60}log\frac{13.1-(-3.8)}{11.6-\left(3.8\right)}=0.001549$
For 120 minutes
$k=\frac{2.303}{120}log\frac{13.1-(-3.8)}{10.2-\left(3.8\right)}=0.001569$
For 180 minutes
$k=\frac{2.303}{180}log\frac{13.1-(-3.8)}{9.0-\left(3.8\right)}=0.001544$
For 360 minutes
$k=\frac{2.303}{360}log\frac{13.1-(-3.8)}{5.87-\left(3.8\right)}=0.001551$
Since, the value of k is constant, the reaction follows first order reaction.
The average value of k is $\frac{0.001549+0.001569+0.001544+0.001551}{4}=\frac{0.0062135}{4}=0.001553mi{n}^{-1}$
To determine the total time, substitute $r_t=0$ in the above expression.
$t=\frac{2.303}{k}log\frac{r_0-r_{\infty }}{r_t-r_{\infty }}$
$t=\frac{2.303}{0.001553}log\frac{13.1-(-3.8)}{0-(-3.8)}=966min$