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Q.

An organic compound undergoes first order decomposition. The time taken for the decomposition to 1/4 and 1/10 of its initial concentration are  t1/4 and  t1/10  respectively. What is the value of  [t1/4][t1/10]×20? (take  log102=0.3)

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a

12

b

6

c

24

d

48

answer is A.

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Detailed Solution

When  t=t114,a=a0/4t1/4=2.303Kloga0a0/4        When   t=t110,a=a0/10   then  t110=2.303Kloga0a0/10              t14t1/10×20=2.303Klog4×K2.303×log10×20        =log4log10×20=2log2log10×20 =2×0.3×201=12

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