Question

$A⟶B$ first order reaction A is optical active and B is optically inactive, a series of experiment were conducted on a solution of A

Time	0	60 min	$\infty$
optical rotation	$82$	$77$	$2$

Assume some impurity present calculate the optical rotation after 5 hours.
$(Givenin1.066=0.064,e^{0.16}=1.17)$

• A. 60
• B. 30
• C. 20
• D. 120

Answer 1

The correct option is A 60

$kt=ln\frac{{\alpha }_0}{{\alpha }_t};k\times 60=ln\frac{80}{75}$

$\Rightarrow k=\frac{1}{60}ln\frac{80}{75}$

$k\times 5\times 60=ln\frac{80}{r_t};300\times \frac{1}{60}.$

$ln\frac{80}{75}=ln\frac{80}{r_t};ln\frac{80}{r_t}=0.32$

$r_t=58$

optical rotation observed after $5$ hrs
$=58+2=60$ {Option A}