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An organic compound undergoes first deco...

An organic compound undergoes first decompoistion. The time taken for its decompoistion to `1//8` and `1//10` of its initial concentration are `t_(1//8)` and `t_(1//10)`, respectively. What is the value of `([t_(1//8)])/([t_(1//10)]) xx 10`? `(log_(10)2 = 0.3)`

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The correct Answer is:
9
For a first order reaction,
`k=(2*303)/t"log"C_0/C_t" or "t=(2*303)/k"log"C_0/C_t`
`:." "1_(1//8)=(2*303)/t"log"C_0/(C_0//8)=(2*303)/k"log 8"`
`=(2*303)/k"log 2"^3=(2*303xx3)/k"log 2"`
`=(2*303xx3xx0*3)/k`
`t_(1//2)=(2*303)/k"log"C_0/(C_0//10)=(2*303)/k" log 10"=(2*303)/k`
`:." "t_(1//8)/t_(1//10)=3xx0*3=0*9" or "t_(1//8)/(t_(1//10))" or "t_(1//8)/t_(1//10)xx10=9`
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Knowledge Check

  • An organic compound undergoes first order decomposition . The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are t_(1//8) and t_(1//10) respectively. What is the value of ([t_(1//8)])/([t_(1//10)]) ? (take log_(10)2=0.3 )

    A
    0.6
    B
    3
    C
    0.9
    D
    9

  • The rate of a first order reaction is 0.6932 times 10^–2 mol ^l–1 min^–1 and the initial concentration of the reactant is 0.1M, then t_(1//2) is equal to

    A
    `0.6932 times 10^-2 min`
    B
    `0.6932 times 10^-3 `min
    C
    10min
    D
    6932min

  • The rate constant of a first order reaction is 6.9 xx 10^(-3)s^(-1) . How much time will it take to reduce the initial concentration to its 1//8^(th) value ?

    A
    100 s
    B
    200 s
    C
    300 s
    D
    400 s
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    1. An organic compound undergoes first decompoistion. The time taken for ...

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