Dear Student, The following information is given: Half life = 150 s Initial concentration = 0 1 M Final concentration = 0 01 M For a first order reactions:k = 0 693t12=0 693150=0 00462 s-1Now,lnA = -kt + lnAo , A = final concentration, Ao=Initial concentrationt = lnAo-lnAk=ln0 1-ln0 010 0462=498 395≈500 Thus, it takes 500 seconds Thus, option (c) is the correct answer Hope it helps Regards

CORRECT OPTION?
CORRECT OPTION? Bookmark Question 28 of 90 a first order reaction with half life of 160 seconds, the time taken for the concentration of the reactant to fall from M/ 10 to M/ 100 will be approximately O 1600 s 900 s 500 s 600 s

Dear Student,

The following information is given:

Half life = 150 s
Initial concentration = 0.1 M
Final concentration = 0.01 M

F o r a f i r s t o r d e r r e a c t i o n s : k = \frac{0.693}{t_{\frac{1}{2}}} = \frac{0.693}{150} = 0.00462 s^{- 1} N o w, \ln A = - k t + \ln A_{o}, A = f i n a l c o n c e n t r a t i o n, A_{o} = I n i t i a l c o n c e n t r a t i o n t = \frac{\ln A_{o} - \ln A}{k} = \frac{\ln 0.1 - \ln 0.01}{0.0462} = 498.395 \approx 500

Thus, it takes 500 seconds.

Thus, option (c) is the correct answer

Hope it helps.
Regards

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​CORRECT OPTION?​CORRECT OPTION? Bookmark Question 28 of 90 a first order reaction with half life of 160 seconds, the time taken for the concentration of the reactant to fall from M/ 10 to M/ 100 will be approximately O 1600 s 900 s 500 s 600 s

CORRECT OPTION?
CORRECT OPTION? Bookmark Question 28 of 90 a first order reaction with half life of 160 seconds, the time taken for the concentration of the reactant to fall from M/ 10 to M/ 100 will be approximately O 1600 s 900 s 500 s 600 s