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An organic compound undergoes first-order decomposition. The t

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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. The value of $\frac{[ t _{1/8} ]}{[ t _{1/10} ]} \times 10$ is ______ (take $l o g_{10} 2 = 0.3$ )

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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. The value of

\frac{[ t _{1/8} ]}{[ t _{1/10} ]} \times 10

is ______ (take

l o g_{10} 2 = 0.3

)

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Text solutionVerified

t = \frac{2.303}{k} l o g (\frac{a}{a - x})

For decomposition to 1/8 of its initial concentration

t_{1/8} = \frac{2.303}{k} l o g (\frac{1}{1/8}) = \frac{2.08}{k}

For decomposition to 1/10 of its initial concentration

t_{1/10} = \frac{2.303}{k} l o g (\frac{1}{1/10}) = \frac{2.303}{k}

Hence,

\frac{[ t _{1/8} ]}{[ t _{1/10} ]} \times 10 = \frac{\frac{2.08}{k}}{\frac{2.303}{k}} \times 10 = 9

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Question Text	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. The value of $\frac{[ t _{1/8} ]}{[ t _{1/10} ]} \times 10$ is ______ (take $l o g_{10} 2 = 0.3$ )
Answer Type	Text solution:1
Upvotes	150