The following first order reaction –
$2\text{ A\hspace{0.17em}(}aq.\text{)}\to 3\text{ B(}aq.\text{)}+4\text{C(}aq.\text{)}$
is monitored by measuring optical rotation of reaction mixture containing optical active dextrorotatory inert impurity at different time interval. The species A, B and C are optically active with specific rotation  ${20}^{\circ },{40}^{\circ }$  and $-{80}^{\circ }$  per mol respectively.

Time (min)	0	22	$\infty$
Optical rotation of reaction mixture	${\mathbf{120}}^o$	$-{\mathbf{40}}^o$	$-{\mathbf{120}}^o$

 (Given  $-\text{ln}3=1.1,\text{ln}2=0.7$ )
which is/are correct statements :

 $2\text{ A}(aq.)\to 3\text{ B}(aq.)+4\text{C}\left(aq\right)$
            
$a$             $0$           $0$         $\text{t}=0$
$a-x$      $\frac{3x}{2}$      $2x$      $\text{t}=22min$
$0$              $\frac{3a}{2}$      $2a$      $t=\infty$
  $\text{a}\times 20=120-\theta$
  $1.5\times 40+2\text{a}\times (-80)=-120-\theta$
  $20a-60a+160a=240$
  $120\text{a}=240\Rightarrow \text{a}=2$
  $\theta =80$
  $(\text{a}-\text{x})\times 20+1.5\text{x}\times 40+2\text{x}(-80)=-40-\theta$
  $40-20\text{x}+60\text{x}-160\text{x}=-40-\theta$
$-120\text{x}=-80-80=-160$
$\text{x}=\frac{160}{120}=\frac{4}{3}$
${\text{ K}}_{\text{A}}=\frac{1}{2}\text{ln}\frac{2}{2-\frac{4}{3}}=\frac{\text{ln}3}{22}=\frac{1.1}{22}$

${\text{ K}}_{\text{A}}=0.05{\min }^{-1};{\text{K}}_{\text{C}}=0.1{\text{ min}}^{-1}$
Time in which $50\text{\%}$  of A.
Reacted  $=\frac{\text{ln}2}{{\text{ K}}_{\text{A}}}=\frac{0.7}{0.05}=14\text{ min}$
Time in which  $75\text{\%}$ of A reacted is $=2\times 14=28\text{ min}$ .