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A first order reaction has the rate constant of $1.15 \times 10^{-3} \mathrm{~s}^{-1}$. The time required to reduce $10 \mathrm{~g}$ of reactant to $6 \mathrm{~g}$ is $\times \times 10^2 \mathrm{sec}$. What is the approximate value of $x$ ?
$(\log 5=0.7, \log 3=0.48)$
Options:
$(\log 5=0.7, \log 3=0.48)$
Solution:
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Verified Answer
The correct answer is:
$4.4$
Order $=1, \mathrm{k}=1.15 \times 10^{-3} \mathrm{~S}^{-1}, \mathrm{C}_1=10, \mathrm{C}_2=6$
$\begin{aligned} & \Rightarrow \mathrm{t}=\frac{2.303}{\mathrm{k}} \log \frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{2.303}{1.15 \times 10^{-3}} \log \frac{10}{6} \\ & =444.27 \cong 4.4 \times 10^2 \\ & \text { Thus, } x=4.4 \\ & \end{aligned}$
$\begin{aligned} & \Rightarrow \mathrm{t}=\frac{2.303}{\mathrm{k}} \log \frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{2.303}{1.15 \times 10^{-3}} \log \frac{10}{6} \\ & =444.27 \cong 4.4 \times 10^2 \\ & \text { Thus, } x=4.4 \\ & \end{aligned}$
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