You visited us 20 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Millikan's Oil Drop Experiment

The reaction,...

Question

The reaction, $S u c r o s e H^{+} - - \to G l u c o s e + F r u c t o s e$ , takes place at certain temperature while the volume of solution is maintained at $1$ litre. At time zero the initial rotation of the mixture is $34^{o} C$ .After $30$ minutes the total rotation of solution is $19^{o} C$ and after a very long time, the total rotation is ${- 11}^{o} C$ . Find the time when solution was optically inactive?

135

min

No worries! We‘ve got your back. Try BYJU‘S free classes today!

103.7

min

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

38.7

min

No worries! We‘ve got your back. Try BYJU‘S free classes today!

45

min

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $103.7$ min
rate constant $k = \frac{2.303}{t} l o g \frac{(r_{0} - r_{\infty})}{(r_{t} - r_{\infty})} = 0.0135$
At the point of optical inactiveness, rotation is zero.
So, time taken is $t = \frac{2.303}{k} l o g (45 / 11) = 103.7$ min.

Suggest Corrections

Similar questions

Q. The reaction,

S u c r o s e \to F r u c t o s e + G l u c o s e

, takes place at a certain temperature while the volume of the solution is maintained at 1 L. At t = 0, the initial rotation of the mixture is

34^{\circ}

. After 30 minutes the total rotation is

19^{\circ}

and after a very long time, the total rotation is

- 11^{\circ}

. Find the time when solution was optically inactive.

Q. The reaction,

S u c r o s e \to F r u c t o s e + G l u c o s e

, takes place at a certain temperature while the volume of the solution is maintained at 1 L. At t = 0, the initial rotation of the mixture is

34^{\circ}

. After 30 minutes the total rotation is

19^{\circ}

and after a very long time, the total rotation is

- 11^{\circ}

. Find the time when solution was optically inactive.

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.

(G i v e n i n 1.066 = 0.064, e^{0.16} = 1.17)

Q. In a hypothetical reaction,

A (a q) ⇌ 2 B (a q) + C (a q)

^{s t}

order decomposition)
'A' is optically active (dextro-rotatory) while 'B' and 'C' are optically inactive but 'B' takes part in a titration reaction (fast reaction) with

H_{2} O_{2}

. Hence the progress of reaction can be monitored by measuring rotation of plane of polarised light or by measuring volume of

H_{2} O_{2}

consumed in titration.

In an experiment, the optical rotation was found to be

θ = 30^{o}

t = 20

min. and

θ = 15^{o}

t = 50

min. from start of the reaction. If the progress would have been monitored by titration method, volume of

H_{2} O_{2}

consumed at

t = 30

. If the progress would have been monitored by titration method, volume of

H_{2} O_{2}

consumed at

t = 30

min. (from start) is 30 ml then volume of

H_{2} O_{2}

consumed at

t = 90

min. will be:

Q. An optically active compound

A

undergoes fist order conversion to another optically active compounds

B

and

C

. From the following data of optical rotation, calculate the time from the start at which the solution will become a racemic mixture.

A \to B + C

time (min)	0	10	$\infty$
optical rotation $(^{\circ})$	$15^{\circ}$	$5^{\circ}$	$- 5^{\circ}$

Join BYJU'S Learning Program

The correct option is D 103.7 minrate constant k=2.303tlog(r0−r∞)(rt−r∞)=0.0135At the point of optical inactiveness, rotation is zero. So, time taken is t=2.303klog(45/11)=103.7 min.

The correct option is D $103.7$ min
rate constant $k = \frac{2.303}{t} l o g \frac{(r_{0} - r_{\infty})}{(r_{t} - r_{\infty})} = 0.0135$
At the point of optical inactiveness, rotation is zero.
So, time taken is $t = \frac{2.303}{k} l o g (45 / 11) = 103.7$ min.