For a first order reaction, the time required for completion of 90...

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 $\log 2 = 0 . 3010$ )

Options

A1.12
B2.43
C3.32
D33.31

Correct answer

C. 3.32

Step-by-step solution

$\begin{array}{ll}\left[\dfrac{\ln 2}{{t}_{1/2}}\right]{t}_{90%} = \ln \left[\dfrac{100}{10}\right] = \ln 10 \\ {t}_{90%} \times \dfrac{0 .693}{{t}_{1/2}} = 2 .303 \\ {t}_{90%} = \dfrac{2 . 303}{0693} \times {t}_{1/2} = 3 .32 {t}_{1/2}\end{array}$

Hence, $x = 3 . 32$

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