For a first order reaction, the time required for completion of 90% reaction is 'x' times the half life of the reaction. The value of 'x' is (Given: ln 10 = 2.303 and log 2 = 0.3010)

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2.43

1.12

3.32

33.31

answer is C.

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Detailed Solution

$Given t_{0 .90} = {xt}_{1 / 2} First order rate constant \begin{aligned} K & = \frac{\ln 2}{t_{1 / 2}} = \frac{1}{{xt}_{1 / 2}} \ln \frac{A_{0}}{A_{0} - A_{0} \times \frac{90}{100}} \\ \frac{\ln 2}{t_{1 / 2}} = \frac{\ln 10}{{xt}_{1 / 2}} \\ x & = \frac{\ln 10}{\ln 2} = \frac{2 .303}{2 .303 \times 0 .3010} = 3 .32 \end{aligned}$

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