In a hypothetical reaction A(aq)hArr2B(aq)+C(aq) (1^(st) order decomposition) A(aq)hArr2B(aq)+C(aq) (1^(st) order decomposition) 'A' is optically active (dextro-rototory) while 'B' and 'C' are optically inactive but 'B' takes part in a titration reaction (fast reaction) with H_2O_2.Hence the progress of reaction can be monitored by measuring rotation of plane of polarised light or by measuring volume of H_2O_2 consumed in titration. In an experiment, the optical rotation was found to be theta=30^@C at t=20 min. and theta=15^(@) at t=50min from start of the reaction.If the progress would have been monitored by titration method, volume of H_2O_2 consumed at t=30 min.(from start) is 30 ml then volume of H_2O_2 consumed at t=90 min, will be:

In a hypothetical reaction A(aq)hArr2B(aq)+C(aq) (1^(st) order decomp

1.	In a hypothetical reaction A(aq)hArr2B(aq)+C(aq) (1^(st) order decomposition) A(aq)hArr2B(aq)+C(aq) (1^(st) order decomposition) 'A' is optically active (dextro-rototory) while 'B' and 'C' are optically inactive but 'B' takes part in a titration reaction (fast reaction) with H_2O_2.Hence the progress of reaction can be monitored by measuring rotation of plane of polarised light or by measuring volume of H_2O_2 consumed in titration. In an experiment, the optical rotation was found to be theta=30^@C at t=20 min. and theta=15^(@) at t=50min from start of the reaction.If the progress would have been monitored by titration method, volume of H_2O_2 consumed at t=30 min.(from start) is 30 ml then volume of H_2O_2 consumed at t=90 min, will be:
Answer» 60 ml 45 ml 52.5 ml 90 ml Solution :As only A is optically active.So conc. Of A at t `=20 "min" PROP 30^@` While CONCENTRATION of A at t`=50 "min" prop 15^@` So concentration has decreased to half of its value in 30 min, so `t_(1//2)`, is according to 50% production of B. At t=90 min, production of B =87.5% (Three half lives) So volume CONSUMED `=(30 ml)+(30/2ml)+(30/4)ml=52.5 ml`